Differential Equations
Derivatives of Polynomials and Exponential Functions
Constant Functions
The derivative of the constant function \(f(x) = c\) is defined as:
$$ f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h} = \lim_{h \rightarrow 0} \frac{c - c}{h} = \lim_{h \rightarrow 0} 0 = 0 $$Derivative of a Constant Function
Power Functions
Let \(f(x) = x^n\), where \(n\) is a positive integer. If \(n = 1\), the graph of \(f(x) = x\) is the line \(y = x\), which has slope \(1\)
Derivative of a Linear Function
For \(n = 4\), we find the derivative of \(f(x) = x^4\) as follows:
$$ f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h} = \lim_{h \rightarrow 0} \frac{(x + h)^4 - x^4}{h} $$$$ = \lim_{h \rightarrow 0} \frac{x^4 + 4x^3h +6x^2h^2 + 4xh^3 + h^4 - x^4}{h} $$$$ = \lim_{h \rightarrow 0} \frac{4x^3h +6x^2h^2 + 4xh^3 + h^4}{h} $$$$ = \lim_{h \rightarrow 0} 4x^3 +6x^2h + 4xh^2 + h^3 = 4x^3 $$We see a pattern emerging.
The Power Rule
First Proof We know that:
$$ x^n - a^n = (x - a)(x^{n - 1} + x^{n - 2}a + \cdots + xa^{n - 2} + a^{n - 1}) $$If \(f(x) = x^n\), then:
$$ f'(a) = \lim_{x \rightarrow a} \frac{f(x) - f(a)}{x - a} = \lim_{x \rightarrow a} \frac{x^n - a^n}{x - a} $$$$ = \lim_{x \rightarrow a} \frac{(x - a)(x^{n - 1} + x^{n - 2}a + \cdots + xa^{n - 2} + a^{n - 1})}{x - a} $$$$ = \lim_{x \rightarrow a} (x^{n - 1} + x^{n - 2}a + \cdots + xa^{n - 2} + a^{n - 1}) $$$$ = a^{n - 1} + a^{n - 2}a + \cdots + a a^{n - 2} + a^{n - 1} $$$$ = na^{n - 1} $$Second Proof
$$ f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h} = \lim_{h \rightarrow 0} \frac{(x + h)^n - x^n}{h} $$We expand \((x + h)^n\) using the Binomial Theorem:
$$ f'(x) = \lim_{h \rightarrow 0} \frac{\left[x^n + nx^{n - 1}h + \frac{n(n - 1)}{2}x^{n - 2}h^2 + \cdots + nxh^{n - 1} + h^n \right] - x^n}{h} $$$$ = \lim_{h \rightarrow 0} \frac{nx^{n - 1}h + \frac{n(n - 1)}{2}x^{n - 2}h^2 + \cdots + nxh^{n - 1} + h^n}{h} $$$$ = \lim_{h \rightarrow 0} nx^{n - 1} + \frac{n(n - 1)}{2}x^{n - 2}h^ + \cdots + nxh^{n - 2} + h^{n - 1} = nx^{n - 1} $$because every term except the first has \(h\) as a factor and therefore approaches \(0\).
But, what about power functions with negative integer exponents? See for example the derivative for \(\frac{1}{x}\):
$$ \frac{d}{dx} \frac{1}{x} = -\frac{1}{x^2} $$We can rewrite these functions as follows:
$$ \frac{d}{dx} (x^{-1}) = (- 1) x^{-2} $$and so the Power Rule is true when \(n = -1\).
What if the exponent is a fraction:
$$ \frac{d}{dx} \sqrt{x} = \frac{1}{2\sqrt{x}} $$which can be rewritten as
$$ \frac{d}{dx} (x^{\frac{1}{2}}) = \frac{1}{2}x^{-\frac{1}{2}} $$The Power Rule (General Version)
If \(n\) is any real nmber, then
$$ \frac{d}{dx} x^n = nx^{n - 1} $$In general, a function increases when its derivative is positive and decreases when it derivative is negative.
The Power Rule enables us to find normal lines. The normal line to a curve \(C\) at a point \(P\) is the line through \(P\) that is perpendicular to the tangent line at \(P\).
New Derivatives from Old
The Constant Multiple Rule
If \(c\) is a constant and \(f\) is a differentiable function, then
$$ \frac{d}{dx} [c(fx)] = c \frac{d}{dx} f(x) $$PROOF Let \(g(x) = cf(x)\). Then
$$ g'(x) = \lim_{h \rightarrow 0} \frac{g(x + h) - g(x)}{h} = \lim_{h \rightarrow 0} \frac{cf(x + h) - cf(x)}{h} $$$$ = \lim_{h \rightarrow 0} c\left[\frac{f(x + h) - f(x)}{h}\right] $$By the Limit Laws:
$$ = c \lim_{h \rightarrow 0}\left[\frac{f(x + h) - f(x)}{h}\right] = cf'(x) $$Sum and Difference Rules
If \(f\) and \(g\) are both differentiable, then
$$ \frac{d}{dx} [f(x) + g(x)] = \frac{d}{dx} f(x) + \frac{d}{dx} g(x) $$$$ \frac{d}{dx} [f(x) - g(x)] = \frac{d}{dx} f(x) - \frac{d}{dx} g(x) $$Proof To prove the Sum Rule, we let \(F(x) = f(x) + g(x)\). Then:
$$ F'(x) = \lim_{h \rightarrow 0} \frac{F(x + h) - F(x)}{h} $$$$ = \lim_{h \rightarrow 0} \frac{[f(x + h) + g(x + h)] - [f(x) + g(x)]}{h} $$$$ = \lim_{h \rightarrow 0} \left[\frac{f(x + h) - f(x)}{h} + \frac{g(x + h) - g(x)]}{h} \right] $$By the Limit Laws:
$$ = \left[\lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h}\right] + \left[\lim_{h \rightarrow 0}\frac{g(x + h) - g(x)]}{h} \right] $$$$ = f'(x) + g'(x) $$To prove the Difference Rule, we write \(f - g\) as \(f + (-1)g\) and apply the Sum Rule and the Constant Multiple Rule.
The Sum Rule can be extended to the sum of any number of functions. For instance, using this theorem twice, we get
$$ (f + g + h)' = [(f + g) + h]' = (f + g)' + h' = f' + g' + h' $$The Constant Multiple Rule, the Sum Rule, and the Difference Rule can be combined with the Power Rule to differentiate any polynomial.
Let \(p(x) = a*n x^n + a*{n - 1}x^{n-1} + \cdots + a_1 x + a_0\), then:
$$ p'(x) = \frac{d}{dx}(a_n x^n + a_{n - 1}x^{n-1} + \cdots + a_1 x + a_0) $$By the Sum Rule and the Constant Multiple Rule:
$$ = \sum_{i=0}^n \frac{d}{dx} a_i x^i = \sum_{i=0}^n a_i \frac{d}{dx} x^i $$By the definition for the derivative of a power function:
$$ = \sum_{i=0}^n (a_i)(i)x^{i-1} $$Example
$$ \frac{d}{dx} (x^8 + 12x^5 - 4x^4 + 10x^3 - 6x + 5) $$$$ = \frac{d}{dx} (x^8) + \frac{d}{dx} (12x^5) + \frac{d}{dx} (-4x^4) + \frac{d}{dx} (10x^3) + \frac{d}{dx} (-6x) + \frac{d}{dx} (5) $$$$ = 8x^7 + (12)(5)x^4 + (-4)(4)x^3 + (10)(3)x^2 + -6 + 0 $$Exponential Functions
Let’s try to compute the derivative of the exponential function \(f(x) = b^x\) using the definition of a derivative:
$$ f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h} = \lim_{h \rightarrow 0} \frac{b^{x + h} - b^x}{h} $$$$ = \lim_{h \rightarrow 0} \frac{b^xb^h - b^x}{h} = \lim_{h \rightarrow 0} \frac{b^x(b^h - 1)}{1} $$The factor \(b^x\) does not depend on \(h\), so we can take it in front of the limit:
$$ = b^x \lim_{h \rightarrow 0} \frac{b^h - 1}{h} $$Notice that the limit is the value of the derivative of \(f\) at \(0\), that is:
$$ \lim_{h \rightarrow 0} \frac{b^h - 1}{h} = f'(0) $$Therefore we have shown that if the exponential function \(f(x) = b^x\) is differentiable at \(0\), then it is differentiable everywhere and:
$$ f'(x) = f'(0)b^x $$This equation says that the rate of change of any exponential function is proportional to the function itself.
Number \(e\)
\(e\) is the number such that:
$$ \lim{h \to 0} \frac{e^h - 1}{h} = 1 $$Derivative of the Natural Exponential Function
Thus the exponential function \(f(x) = e^x\) has the property that it is its own derivative. The geometrical significance of this fact is that the slope of a tangent line to the curve \(y = e^x\) at a point \((x, e^x)\) is equal to the \(y\)-coordinate of the point.
The Product and Quotient Rules
The Product Rule
The Product Rule
If \(f\) and \(g\) are both differentiable, then
$$\frac{d}{dx}[f(x)g(x)] = f(x)\frac{d}{dx}[g(x)] + g(x) \frac{d}{dx}[f(x)]$$The Product Rule’s Proof 1
By the definition of a derivative the derivative of any function $f$ is given by:
$$ \frac{\delta}{\delta x} f(x) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} $$Given two functions $f$ and $g$, we use the definition of the derivative for the product function $F = (f \cdot g)$:
$$ \frac{\delta}{\delta x} F(x) = \lim_{\Delta x \to 0} \frac{F(x + \Delta x) - F(x)}{\Delta x} $$$$ = \lim_{\Delta x \to 0} \frac{f(x + \Delta x)g(x + \Delta x) - f(x)g(x)}{\Delta x} $$We sum and substract $f(x + \Delta x)g(x)$
$$ = \lim_{\Delta x \to 0} [\frac{f(x + \Delta x)g(x + \Delta x) - f(x + \Delta x)g(x)] + [f(x + \Delta x)g(x) - f(x)g(x)]}{\Delta x} $$$$ = \lim_{\Delta x \to 0} \left[\frac{f(x + \Delta x)g(x + \Delta x) - f(x + \Delta x)g(x)}{\Delta x}\right] + \left[\frac{f(x + \Delta x)g(x) - f(x)g(x)}{\Delta x}\right] $$$$ = \lim_{\Delta x \to 0} \left[f(x + \Delta x)\frac{g(x + \Delta x) - g(x)}{\Delta x}\right] + \lim_{\Delta x \to 0}\left[g(x)\frac{f(x + \Delta x) - f(x)}{\Delta x}\right] $$$$ = \left[\lim_{\Delta x \to 0}f(x + \Delta x) \lim_{\Delta x \to 0} \frac{g(x + \Delta x) - g(x)}{\Delta x}\right] + \left[\lim_{\Delta x \to 0} g(x) \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}\right] $$Given $\lim_{\Delta x \to 0} f(x + \Delta x) = f(x)$, then
$$ = f(x) \frac{\delta g(x)}{\delta x} + g(x) \frac{\delta f(x)}{\delta x} $$The Product Rule’s Proof 2
We start by assuming that \(u = f(x)\) and \(v = g(x)\) are both positive differentiable functions. Then we can interpret the product \(uv\) as an area of a rectangle (see Figure 1).
If \(x\) changes by and amount \(\Delta x\), then de corresponding changes in \(u\) and \(v\) are
$$ \Delta u = f(x + \Delta x) - f(x) $$$$ \Delta v = g(x + \Delta x) - g(x) $$The change in the area of the rectangle is
$$ \Delta(uv) = (u + \Delta u)(v + \Delta v) - uv = u\Delta v + v \Delta u + \Delta u \Delta v $$where \((u + \Delta u)(v + \Delta v)\) can be interpreted as the area of the large rectangle in Figure 1. Whilst \(\Delta(uv)\) would be the sum of the tree shaded areas.
If we divide by \(\Delta x\), we get:
$$ \frac{\Delta(uv)}{\Delta x} = u \frac{\Delta v}{\Delta x} + v \frac{\Delta u}{\Delta x} + \frac{\Delta u \Delta v}{\Delta x} $$If we now let \(\Delta x \rightarrow 0\) we get the derivative of \(uv\):
$$ \frac{\delta}{\delta x}(uv) = \lim_{\Delta x \to 0} \frac{\Delta (uv)}{\Delta x} $$$$ = \lim_{\Delta x \to 0} \left(u \frac{\Delta v}{\Delta x} + v \frac{\Delta u}{\Delta x} + \frac{\Delta u \Delta v}{\Delta x}\right) $$$$ = u \lim_{\Delta x \to 0} \frac{\Delta v}{\Delta x} + v \lim_{\Delta x \to 0} \frac{\Delta u}{\Delta x} + \left(\lim_{\Delta x \to 0} \Delta u \right)\left(\lim_{\Delta x \to 0} \frac{ \Delta v}{\Delta x}\right) $$Substituting by \(\Delta v = g(x + \Delta x) - g(x)\) and \(\Delta u = f(x + \Delta x) - f(x)\)
$$ = u \lim_{\Delta x \to 0} \frac{g(x + \Delta x) - g(x)}{\Delta x} + v \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} + \left(\lim_{\Delta x \to 0} \Delta u \right)\left(\lim_{\Delta x \to 0} \frac{ \Delta v}{\Delta x}\right) $$Because $\lim_{\Delta x \to 0} \Delta u = 0$
$$ = u \lim_{\Delta x \to 0} \frac{g(x + \Delta x) - g(x)}{\Delta x} + v \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} + 0 \cdot \left(\lim_{\Delta x \to 0} \frac{ \Delta v}{\Delta x}\right) $$$$ = u \lim_{\Delta x \to 0} \frac{g(x + \Delta x) - g(x)}{\Delta x} + v \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} $$Although we started by assuming (for the geometric interpretation) that all the quantities are positive, we notice that The Power Rule is always true. (The algebra is valid whether $u$, $v$, $\Delta u$, and $\Delta v$ are positive or negative).
The Quotient Rule
The Quotient Rule
If \(f\) and \(g\) are both differentiable, then
$$\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{g(x)\frac{d}{dx}[f(x)] - f(x) \frac{d}{dx}[g(x)]}{[g(x)]^2}$$The Quotient Rule’s Proof 1
Given the quotient function $F = \frac{f}{g}$, then by the definition of a derivative
$$ \frac{\delta F(x)}{\delta x} = \lim_{\Delta x \to 0} \frac{F(x + \Delta x) - F(x)}{\Delta x} $$$$ = \lim_{\Delta x \to 0} \frac{\frac{f(x + \Delta x)}{g(x + \Delta x)} - \frac{f(x)}{g(x)}}{\Delta x} $$$$ = \lim_{\Delta x \to 0} \frac{\frac{g(x)f(x + \Delta x) - f(x)g(x + \Delta x)}{g(x) \cdot g(x + \Delta x)}}{\Delta x} $$$$ = \lim_{\Delta x \to 0} \frac{g(x)f(x + \Delta x) - f(x)g(x + \Delta x)}{\Delta x \cdot g(x) \cdot g(x + \Delta x)} $$We sum and substract $f(x)g(x)$ on the numerator
$$ = \lim_{\Delta x \to 0} \frac{[g(x)f(x + \Delta x) - f(x)g(x)] + [f(x)g(x) - f(x)g(x + \Delta x)]}{\Delta x \cdot g(x) \cdot g(x + \Delta x)} $$$$ = \lim_{\Delta x \to 0} \frac{g(x)[f(x + \Delta x) - f(x)] + f(x)[g(x) - g(x + \Delta x)]}{\Delta x \cdot g(x) \cdot g(x + \Delta x)} $$$$ = \lim_{\Delta x \to 0} \frac{g(x)[f(x + \Delta x) - f(x)] - f(x)[g(x + \Delta x) - g(x)]}{\Delta x \cdot g(x) \cdot g(x + \Delta x)} $$$$ = \frac{\lim_{\Delta x \to 0} \frac{g(x)[f(x + \Delta x) - f(x)]}{\Delta x} - \lim_{\Delta x \to 0}\frac{f(x)[g(x + \Delta x) - g(x)]}{\Delta x}}{\lim_{\Delta x \to 0} g(x) \cdot g(x + \Delta x)} $$$$ = \frac{g(x)\lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} - f(x)\lim_{\Delta x \to 0}\frac{g(x + \Delta x) - g(x)}{\Delta x}}{\lim_{\Delta x \to 0} g(x) \cdot g(x + \Delta x)} $$Given $\lim_{\Delta x \to 0} g(x + \Delta x) = 0$
$$ = \frac{g(x) \frac{\delta f(x)}{\delta x} - f(x)\frac{\delta g(x)}{\delta x}}{(g(x))^2} $$The Quotient Rule’s Proof 2
If $x$, $u$, and $v$ change by amounts $\Delta x$, $\Delta u$ and $\Delta v$, the corresponding change in the quotient $\frac{u}{v}$ is
$$ \Delta \left(\frac{u + \Delta u}{v + \Delta v}\right) - \frac{u}{v} = $$$$ = \frac{(u + \Delta u)v - u(v + \Delta v)}{v(v + \Delta v)} $$$$ = \frac{uv + v \Delta u - uv - u\Delta v}{v(v + \Delta v)} $$$$ = \frac{v \Delta u - u\Delta v}{v(v + \Delta v)} $$Therefore
$$ \frac{d}{dx} \left(\frac{u}{v}\right) = \lim_{\Delta x \to 0} \frac{\Delta \frac{u}{v}}{\Delta x} = \lim_{\Delta x \to 0} \frac{\frac{v \Delta u - u\Delta v}{v(v + \Delta v)}}{\Delta x} $$$$ = \lim_{\Delta x \to 0} \frac{\frac{v \Delta u - u\Delta v}{\Delta x}}{v(v + \Delta v)} $$$$ =\lim_{\Delta x \to 0} \frac{v\frac{\Delta u}{\Delta x} - u\frac{\Delta v}{\Delta x}}{v(v + \Delta v)} $$$$ = \frac{v \lim_{\Delta x \to 0}\frac{\Delta u}{\Delta x} - u \lim_{\Delta x \to 0} \frac{\Delta v}{\Delta x}}{v \lim_{\Delta x \to 0} (v + \Delta v)} $$Substituting by \(\Delta v = g(x + \Delta x) - g(x)\) and \(\Delta u = f(x + \Delta x) - f(x)\)
$$ = \frac{v \lim_{\Delta x \to 0}\frac{f(x + \Delta x) - f(x)}{\Delta x} - u \lim_{\Delta x \to 0} \frac{g(x + \Delta x) - g(x)}{\Delta x}}{v \lim_{\Delta x \to 0} (v + \Delta v)} $$$$ = \frac{v \frac{d}{dx} f(x) - u \frac{d}{dx} g(x)}{v \lim_{\Delta x \to 0} (v + \Delta v)} $$We know that as $\Delta x \rightarrow 0$ then $\Delta v \rightarrow 0$, therefore $\lim_{\Delta x \to 0} (v + \Delta v) = v + 0 = v$ and
$$ = \frac{v \frac{d}{dx} f(x) - u \frac{d}{dx} g(x)}{v^2} $$Given $u = f(x)$ and $v = g(x)$, then
$$ \frac{d}{dx} \left(\frac{f(x)}{g(x)}\right)= \frac{g(x) \frac{d}{dx} f(x) - f(x) \frac{d}{dx} g(x)}{[g(x)]^2} $$Derivative of Trigonometric Funcions
If we sketch the graph of the function $f(x) = \sin(x)$ and use the interpretation of $f’(x)$ as the slope of the tangent to the sine curve in order to sketch the graph of $f’$ then it looks as if the graph of $f’$ may be the same as the cosine curve (see Figure 1).
From the definition of a derivative, we have
$$ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} $$$$ = \lim_{h \to 0} \frac{\sin(x + h) - \sin x}{h} $$By the Sine Sum and Difference Identities
$$ = \lim_{h \to 0} \left[\frac{\sin(x)\cos(h) + \sin(h)\cos(x) - \sin x}{h}\right] $$$$ = \lim_{h \to 0} \left[\frac{\sin(x)\cos(h) - \sin(x)}{h} + \frac{\sin(h)\cos(x)}{h}\right] $$$$ = \lim_{h \to 0} \left[\sin x \frac{\cos(h) - 1}{h} + \cos x \frac{\sin(h)}{h}\right] $$$$ = \lim_{h \to 0} \sin x \lim_{h \to 0}\frac{\cos(h) - 1}{h} + \lim_{h \to 0}\cos x \lim_{h \to 0}\frac{\sin(h)}{h} $$Because we regard $x$ as a constant when computing a limit as $h \to 0$, we have
$$ \lim_{h \to 0} \sin x = \sin x $$and
$$ \lim_{h \to 0} \cos x = \cos x $$Later we will prove that
$$ \lim_{h \to 0} \frac{\sin h}{h} = 1 $$and
$$ \lim_{h \to 0} \frac{\cos h - 1}{h} = 0 $$If we substitute these limits on our main equation we get
$$ = \sin x \cdot 0 + \cos x 1 = \cos x $$So we have proved the formula for the derivative of the sine function:
$$ \frac{\delta}{\delta x} \sin(x)= \cos x $$Using the same methods as in the previous proof, we can prove
$$ \frac{\delta}{\delta x} \cos(x)= -\sin x $$The tangent function can also be differentiated by using the definition of a derivative, but it is easier to use the Quotient Rule
$$ \frac{\delta}{\delta x} \tan x = \frac{\delta}{\delta x} \left(\frac{\sin x}{\cos x}\right) $$$$ = \frac{\cos x \frac{\delta}{\delta x}\sin x - \sin x \frac{\delta}{\delta x} \cos x}{\cos^2 x} $$$$ = \frac{\cos x \cos x - \sin x \sin x}{\cos^2 x} $$$$ = \frac{\cos^2 x - \sin^2 x}{\cos^2 x} $$Given $\cos^2 x + \sin^2 x = 1$
$$ = \frac{1}{\cos^2 x} = \sec^2 x $$Therefore
$$ \frac{\delta}{\delta x} \tan x = \sec^2 x $$The derivatives of the remaining trigonometric functions, $\csc$, $\sec$, and $\cot$, can also be found easily using the Quotient Rule. We collect all the differentiation formulas for trigonometric functions in the following table. Remember that they are valid only when $x$ is measured in radians.
Derivatives of Trigonometric Functions
Two Special Trigonometric Limits
In proving the formula for the derivative of sine we used two special limits, which we now prove.
$$ \lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 $$Proof for the first fundamental limit
Given a unit circle, asumme first that $\theta$ lies between $0$ and $\frac{\pi}{2}$ (See Figure 6). By the definition of a radian measure
$$ s = arc AB = \theta $$Also
$$ |BC| = |OB| \sin \theta = \sin \theta $$
From the diagram we see that:
$$ |BC| < |AB| < arc AB $$By definition we know that $\sin \theta = \frac{|BC|}{1} = |BC|$, therefore:
$$ \sin \theta = |BC| < arc AB = \theta $$so
$$ \frac{\sin \theta}{\theta} < 1 $$Let the tangent lines at $A$ and $B$ interset at $E$. You can see from Figure 6(b) that the circumference of a circle is smaller than the length of a circumscribed polygon, and so
$$ arc AB < |AE| + |EB| $$Thus
$$ \theta = arc AB < |AE| + |EB| $$From the diagram $|EB| < |ED|$, as $ED$ is the hypothenuse of $BDE$.
$$ < |AE| + |ED| = |AD| $$Also, from the definition of the tangent function
$$ |AD| = |OA| \tan \theta = 1 \tan \theta = \tan \theta $$Therefore we have
$$ \theta < |AD| = \tan \theta = \frac{\sin \theta}{\cos \theta} $$Which we can rewrite as
$$ \cos \theta < \frac{\sin \theta}{\theta} $$so because we also know that $\frac{\sin \theta}{\theta} < 1$, then
$$ \cos \theta < \frac{\sin \theta}{\theta} < 1 $$We know that
$$ \lim_{\theta \to 0} 1 = 1 $$and
$$ \lim_{\theta \to 0} \cos \theta = 1 $$so by the Squeeze Theorem
$$ \lim_{\theta \to 0^+} \frac{\sin \theta}{\theta} = 1, 0 < \theta < \frac{\pi}{2} $$But the function $\frac{\sin \theta}{\theta}$ is an even function, so its right and left limits must be equal. Hence we have
$$ \lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 $$Proof for the second fundamental limit
The following special limit involves cosine.
$$ \lim_{\theta \to 0} \frac{\cos \theta - 1}{\theta} = 0 $$We multiply numerator and denominator by $\cos \theta + 1$
$$ \lim_{\theta \to 0} \frac{\cos \theta - 1}{\theta} = \lim \left(\frac{\cos \theta - 1}{\theta} \cdot \frac{\cos \theta + 1}{\cos \theta + 1}\right) $$$$ = \lim_{\theta \to 0} \frac{\cos^2 \theta - 1}{\theta (\cos \theta + 1)} $$Given $\sin^2 \theta + \cos^2 \theta = 1$
$$ = \lim_{\theta \to 0} \frac{-\sin^2 \theta}{\theta (\cos \theta + 1)} $$$$ = -\lim_{\theta \to 0} \frac{\sin \theta}{\theta} \cdot \frac{\sin \theta}{\cos \theta + 1} $$$$ = -\lim_{\theta \to 0} \frac{\sin \theta}{\theta} \cdot \lim_{\theta \to 0} \frac{\sin \theta}{\cos \theta + 1} $$$$ = -1 \cdot \left(\frac{0}{1 + 1}\right) = 0 $$The Chain Rule
It turns out that the derivative of the composite function $f \circ g$ is the product of the derivatives of $f$ and $g$. This fact is one of the most important of the differentiation rules and is called the Chain Rule.
If we let $y = f(u)$ and $u = g(x)$, then if $u$ changes twice as fast as $x$ and $y$ changes three times as fast as $u$, then it seems reasonable that $y$ changes six times as fast as $x$, and so we expect that $\frac{\delta y}{\delta x}$ is the product of $\frac{\delta y}{\delta u}$ and $\frac{\delta u}{\delta x}$.
The Chain Rule
If $g$ is differentiable at $x$ and $f$ is differentiable at $g(x)$, then the composite function $F= f \circ g$ defined by $F(x) = f(g(x))$ is differentiable at $x$ and $F’$ is given by the product
$$F'(x) = f'(g(x)) \cdot g'(x)$$In Leibniz notation if $y = f(u)$ and $u = g(x)$ are both differentiable functions, then
$$\frac{\delta y}{\delta x} = \frac{\delta y}{\delta u} \frac{\delta u}{\delta x}$$Note that in using the Chain Rule we work from the outside to the inside. We differentiate the outer function $f$ [at the inner function $g(x)$] and then we multiply by the derivative of the inner function.
The Power Rule Combined with the Chain Rule
If $n$ is any real number and $u = g(x)$ is differentiable, then
$$\frac{\delta}{\delta x} (u^n) = nu^{n - 1} \frac{\delta u}{\delta x}$$Alternatively,
$$\frac{\delta}{\delta x} [g(x)]^n = n[g(x)]^{n - 1}g'(x)$$The reason for the name Chain Rule becomes clear when we make a longer chain by adding another link. Suppose that $y = f(u)$, $u = g(x)$, and $x = h(t)$ where $f$, $g$, and $h$ are differentiable functions. Then, to compute the derivative of $y$ with respect to $t$, we use the Chain Rule twice:
$$ \frac{\delta x}{\delta t} = \frac{\delta y}{\delta x} \frac{\delta x}{\delta t} = \frac{\delta y}{\delta u}\frac{\delta u}{\delta x}\frac{\delta x}{\delta t} $$Derivatives of General Exponential Functions
We can use the Chain Rule to differentiate an exponential function with any base $b > 0$. We can write
$$ b^x = \left(e^{\ln b}\right)^x = e^{(\ln b)x} $$and then the Chain Rule gives
$$ \frac{\delta}{\delta x} (b^x) = \frac{\delta }{\delta x} (e^{(\ln b)x}) $$$$ = \frac{\delta e^{(\ln b)x}}{\delta (\ln b)x} \frac{\delta (\ln b)x}{\delta x} $$$$ e^{(\ln b)x} \ln b = b^x \ln b $$So we have the formula
$$ \frac{\delta}{\delta} (b^x) = b^x \ln b $$How to Prove the Chain Rule
Recall that if $y = f(x)$ and $x$ change from $a$ to $a + \Delta x$, we define the increment of $y$ as
$$ \Delta y = f(a + \Delta x) - f(a) $$According to the definition of a derivative we have
$$ \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = f'(a) $$So if we denote the difference between $\frac{\Delta y}{\Delta x}$ and $f’(a)$ by $\epsilon$, we obtain
$$ \lim_{\Delta x \to 0} \epsilon = \lim_{\Delta x \to 0} \left(\frac{\Delta y}{\Delta x} - f'(a)\right) $$$$ = \left(\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}\right) - \left(\lim_{\Delta x \to 0} f'(a)\right)= f'(a) - f'(a) $$But
$$ \epsilon = \frac{\Delta y}{\Delta x} - f'(a) \Rightarrow \Delta y = f'(a)\Delta x + \epsilon \Delta x $$If we define $\epsilon$ to be $0$, when $\Delta x = 0$, then $\epsilon$ becomes a continuous funcion of $\Delta x$ because, as we have seen before
$$ \lim_{\Delta x \to 0} \epsilon = 0 = \epsilon $$Therefore, by the definition of a continuous function we can say that $\epsilon$ is continuous. Thus for a differentiable function $f$, we can write
$$ \Delta y = f'(a)\Delta x + \epsilon \Delta x $$where $\epsilon \to 0$ as $\Delta x \to 0$. And $\epsilon$ is a continuous function of $\Delta x$. This property of differentiable functions is what enables us to prove the Chain Rule.
Chain Rule’s Proof 1
Given the composite function
$$ y = f(g(x)) $$By the definition of a derivative
$$ \frac{\delta y}{\delta x} = \lim_{\Delta x \to 0} \frac{f(g(x + \Delta x)) - f(g(x))}{\Delta x} $$Let’s define the increment on $u$ as
$$ \Delta u = g(x + \Delta x) - g(x) $$Given $g$ is differentiable, when $\Delta x \to 0$, then $\Delta u \to 0$. We rewrite the derivative as
$$ = \lim_{\Delta x \to 0} \frac{f(g(x + \Delta x)) - f(g(x))}{\Delta u} \frac{\Delta u}{\Delta x} $$$$ = \left(\lim_{\Delta x \to 0} \frac{f(g(x + \Delta x)) - f(g(x))}{\Delta u}\right) \left(\lim_{\Delta x \to 0} \frac{\Delta u}{\Delta x} \right) $$The first term is the derivative of $f$ with respecto to $u$ and the second term is the derivative of $u$ with respect to $x$. Given
$$ u = g(x) $$and
$$ \Delta u = g(x + \Delta x) - g(x) \Leftrightarrow g(x + \Delta x) = g(x) + \Delta u = u + \Delta u $$Then
$$ = \left(\lim_{\Delta x \to 0} \frac{f(u + \Delta u) - f(u)}{\Delta u}\right) \left(\lim_{\Delta x \to 0} \frac{g(x + \Delta x) - g(x)}{\Delta x} \right) $$$$ = \frac{\delta f(u)}{\delta u} \frac{\delta g(x)}{\delta x} $$$$ = \frac{\delta f(g(x))}{\delta g(x)} \frac{\delta g(x)}{\delta x} $$Chain Rule’s Proof 2
Suppose $u = g(x)$ is differentiable at $a$ and $y = f(u)$ is differentiable at $b = g(a)$. If $\Delta x$ is an increment in $x$ and $\Delta u$ and $\Delta y$ are the corresponding increments in $u$ and $y$, we previously defined a generic $\Delta y$ as
$$ \Delta y = f'(a)\Delta x + \epsilon \Delta x $$So we can write
$$ \Delta u = g'(a) \Delta x + \epsilon_1 \Delta x = [g'(a) + \epsilon_1] \Delta x $$where $\epsilon_1 \to 0$ as $\Delta x \to 0$. Similarly
$$ \Delta y = f'(b) \Delta u + \epsilon_2 \Delta u = [f'(b) + \epsilon_2] \Delta u $$where $\epsilon_2 \to 0$ as $\Delta u \to 0$.
If we substitute the expression for $\Delta u$ on this last equation
$$ \Delta y = [f'(b) + \epsilon_2] \Delta u = [f'(b) + \epsilon_2] [g'(a) + \epsilon_1] \Delta x $$$$ \frac{\Delta y}{\Delta x} = [f'(b) + \epsilon_2] [g'(a) + \epsilon_1] $$We also know that $\Delta u \to 0$ as $\Delta x \to 0$ because of its defnition:
$$ \Delta u = [g'(a) + \epsilon_1] \Delta x $$So taking the limit as $\Delta x \to 0$, we get
$$ \frac{\delta y}{\delta x} = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = \lim_{\Delta x \to 0} [f'(b)+\epsilon_2][g'(a) + \epsilon_1] $$$$ = \left(\lim_{\Delta x \to 0} [f'(b)+\epsilon_2]\right) \cdot \left(\lim_{\Delta x \to 0} [g'(a) + \epsilon_1]\right) $$By definition $\epsilon_1 \to 0$ and $\epsilon_2 \to 0$ as $\Delta x \to 0$, therefore
$$ = f'(b)g'(a) = f'(g(a))g'(a) $$This proves the Chain Rule.
Implicit Differentiation
Some functions are defined implicitly by a relation between $x$ and $y$, where it is not easy to define the function as dependent only on $x$. For example:
$$ x^2 + y^2 = 25 $$In some cases it is possible to solve such an equation for $y$ as an explicit function (or several functions) of $x$ (See Figure 1.)
$$ y = \pm \sqrt{25 - x^2} $$
Fortunately, we don’t need to solve an equation for $y$ in terms of $x$ in order to find the derivative of $y$. Instead we can use the method of implicit differentiation. This consists of
- Differentiating both sides of the equation with respect to $x$
- Solving the resulting equation for $\frac{\delta y}{\delta x}$.
Preservation of Equality
When we define a function implicitly we are stating that both sides of the equation are always equal. But how do we know that when we differentiate both sides this equality is maintained?.
Geometrically, you can imagine the two sides of the equation as two functions that always overlap. Therefore we could argue that the derivatives of functions that are equivalent must be equal.
We could also prove it using limits, let’s say we have an implicit function as follows:
$$ F(x, y) = G(x, y) $$where both $F$ and $G$ are functions dependent on $x$ and $y$. We must show that:
$$ \frac{\delta}{\delta x} F(x) = \frac{\delta}{\delta x} G(x) $$The derivatives for both functions are defined as:
$$ \frac{\delta}{\delta x} F(x, y) = \lim_{\Delta x \to 0} \frac{F(x + \Delta x, y + \Delta y) - F(x, y)}{\Delta x} $$$$ \frac{\delta}{\delta x} G(x, y) = \lim_{\Delta x \to 0} \frac{G(x + \Delta x, y + \Delta y) - G(x, y)}{\Delta x} $$Note that for a change on $x$ given by $\Delta x$ we assume an implicit change on $y$ given by $\Delta y$. That is $\Delta y$ is the change on $y$ that happens because $y$ depends on $x$.
Given $F(x, y) = G(x, y)$ for all $x$, then for any change $\Delta x$ it follows:
$$ F(x + \Delta x, y + \Delta y) = G(x + \Delta x, y + \Delta y) $$If we substract $F(x, y) = G(x, y)$ on both sides
$$ F(x + \Delta x, y + \Delta y) - F(x, y)= G(x + \Delta x, y + \Delta y) - G(x, y) $$And now we divide both sides by $\Delta x$
$$ \frac{F(x + \Delta x, y + \Delta y) - F(x, y)}{\Delta x}= \frac{G(x + \Delta x, y + \Delta y) - G(x, y)}{\Delta x} $$Therefore, if we take the limit
$$ \lim_{\Delta x \to 0} \frac{F(x + \Delta x, y + \Delta y) - F(x, y)}{\Delta x} = \lim_{\Delta x \to 0}\frac{G(x + \Delta x, y + \Delta y) - G(x, y)}{\Delta x} $$Which therefore proves:
$$ \frac{\delta}{\delta x} F(x, y) = \frac{\delta}{\delta x} G(x, y) $$Example of Implicit Derivation
EXAMPLE
Find $\frac{\delta y}{\delta x}$ for $x^2 + y^2 = 25$
We start by differentiating both sides of the equation
$$\frac{\delta}{\delta x} (x^2 + y^2) = \frac{\delta}{\delta x} 25$$$$\frac{\delta}{\delta x} (x^2) + \frac{\delta}{\delta x}(y^2) = 0$$Remembering that $y$ is a function of $x$, using the chain rule, we have
$$\frac{\delta}{\delta x} (y^2) = \frac{\delta}{\delta y} (y^2) \frac{\delta y}{\delta x} = 2y \frac{\delta y}{\delta x}$$Thus
$$2x + 2y \frac{\delta y}{\delta x} = 0$$Now we solve for $\frac{\delta y}{\delta x}$
$$\frac{\delta y}{\delta x} = - \frac{x}{y}$$
Derivatives of Logarithmic and Inverse Trigonometric Functions
Derivatives of Logarithmic Functions
If $f$ is a one-to-one differentiable function, then its inverse function $f^{-1}$ is also differentiable. Note that if $f$ has a horizontal tangent at a point, then $f^{-1}$ has a vertical tangent at the corresponding reflected point and so $f^{-1}$ is not differentiable there.
We now state and prove the formula for the derivative of a logarithmic function.
$$ \frac{\delta}{\delta x} (\log_b x) = \frac{1}{x \ln b} $$Proof for the Derivative of a Logarithmic Function
Let $y = \log_b x$, then:
$$ b^y = x $$Differentiating this equation implicitly with respecto to $x$ we get
$$ \frac{\delta}{\delta y} b^y \frac{\delta y}{\delta x} = 1 $$$$ b^y \ln b \frac{\delta y}{\delta x} = 1 $$$$ \frac{\delta y}{\delta x} = \frac{1}{b^y \ln b} = \frac{1}{x \ln b} $$Now, if we let $b = e$, then $\ln e = 1$ and so:
$$ \frac{\delta}{\delta x} \ln (x) = \frac{1}{x} $$The simplicity of the previous derivative is one of the main reasons that natural logarithms (logarithms with base $e$) are used in calculus.
In general, if we combine the derivative of $e^x$ with the Chain Rule, then:
$$ \frac{\delta}{\delta x} [\ln g(x)] = \frac{1}{g(x)} \frac{\delta g(x)}{\delta x} $$Logarithmic Differentiation
The calculation of derivatives of complicated functions involving products, quotients, or powers can often be simplified by taking logarithms. This mehod is denoted logarithmic differentiation.
- Take natural logarithms of both sides of an equation $y = f(x)$ and use the Laws of Logarithms to expand the expression.
- Differentiate implicitly with respect to $x$
- Solve the resulting equation for $y’$ and replace $y$ by $f(x)$.
The Number $e$ as a limit
We have shown that if $f(x) = \ln x$, then $\frac{\delta f(x)}{\delta x} = \frac{1}{x}$ and $\frac{f(1)}{\delta x} = 1$. From the definition of a derivative as a limit, we have:
$$ \frac{\delta f(1)}{\delta x} = \lim_{h \to 0} \frac{f(1 + h) - f(1)}{h} $$$$ = \lim_{x \to 0} \frac{f(1 + x) - f(1)}{x} $$$$ = \lim_{x \to 0} \frac{\ln (1 + x) - ln(1)}{x} $$Given $\ln(1) = 0$
$$ = \lim_{x \to 0} \frac{1}{x} \ln (1 + x) $$$$ = \lim_{x \to 0} \ln (1 + x)^{\frac{1}{x}} $$Because $\frac{\delta f(1)}{\delta x} = 1$, then
$$ \frac{\delta f(1)}{\delta x} = \lim_{x \to 0} \ln (1 + x)^{\frac{1}{x}} = 1 $$Then
$$ e = e^1 = e^{\lim_{x \to 0} \ln(1 + x)^{\frac{1}{x}}} $$By the continuity of the exponential function and the continuity properties for composite functions:
$$ f(\lim_{x \to 0} g(x)) = \lim_{x \to 0} f(g(x)) $$where $f(x) = e^x$ and $g(x) = \ln(1 + x)^{\frac{1}{x}}$
$$ e^{\lim_{x \to 0} \ln(1 + x)^{\frac{1}{x}}} = \lim_{x \to 0} e^{\ln(1 + x)^{\frac{1}{x}}} $$$$ = \lim_{x \to 0} (1 + x)^{\frac{1}{x}} $$Derivatives of Inverse Trigonometric Functions
Because the trigonometric functions, with the restricted domains that we used to define their inverses, are one-to-one and differentiable, it follows that the inverse trigonometric functions are also differentiable.
Let’s start with the arcsine function, defined as:
$$ y = \sin^{-1}x \Leftrightarrow \sin y = x \text{ and } -\frac{\pi}{2} \leq y \leq \frac{\pi}{2} $$We differentiate $\sin y = x$ implicitly
$$ \cos(y) \frac{\delta y}{\delta x} = 1 \Leftrightarrow \frac{\delta y}{\delta x} = \frac{1}{\cos(y)} $$We know that $\cos y \geq 0$ because $-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$, so
$$ \cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2} $$$$ \frac{\delta y}{\delta x} = \frac{1}{\cos y} = \frac{1}{\sqrt{1 - x^2}} $$Therefore
$$ \frac{\delta}{\delta x} (\sin^{-1}x) = \frac{1}{\sqrt{1 - x^2}} $$Derivatives of Inverse Trigonimetric Functions
Rates of Change in the Natural and Social Sciences
Example
A current exists whenever electric charges move (See Figure 6).
If $\Delta Q$ is the net charge that passes through the red surface during a time period $\Delta t$, then the average current during this time inverval is
$$ \text{averga current } = \frac{\Delta Q}{\Delta t} = \frac{Q_2 - Q_1}{t_2 - t_1} $$If we take the limite over smaller and smaller time intervals, we obtian what is called current $I$ at a given time $t_1$:
$$ I = \lim_{\Delta t \to 0} \frac{\Delta Q}{\Delta t} = \frac{\delta Q}{\delta t} $$Thus the current is the rate at which charge flows through a surface. It is measure in units of charge per unit time (often coulombs per second, called amperes).
Velocity, density, and current are not the only rates of change that are important in physics. Others include power (the rate at which work is done), the rate of heat flow, temperature gradient (the rate of change of temperature with respect to position), and the rate of decay of a radioactive substance in nuclear physics.
Exponential Growth and Decay
In general, if $y(t)$ is the value of a quantity at time $t$, and if the rate of change of $y$ with respect to $t$ is proportional tos size $y(t)$ at any time, then
$$ \frac{\delta y(t)}{\delta t} = ky(t) $$where $k$ is a constant, called relative growth rate. This is sometimes called the law of natural growth (if $k > 0$) or the law of natural decay (if $k < 0$). It is called a differential equation because it involver an unknown function $y$ and its derivative $\frac{\delta y(t)}{\delta t}$
Also note that any exponential function of the form $y(t) = Ce^{kt}$, where $C$ is a constant, satisfies:
$$ \frac{\delta y(t)}{\delta t} = C(ke^{kt}) = k(Ce^{kt}) = ky(t) $$To see the significance of the constant $C$, we observe that
$$ y(0) = Ce^{k\cdot 0} = C $$Therefore $C$ is the initial value for the function.
Population Growth
In the context of population growth, where $P(t)$ is the size of a population at time $t$, we can write:
$$ \frac{\delta P(t)}{\delta t} = kP(t) \text{ or } \frac{\frac{\delta P(t)}{\delta t}}{P(t)} = k $$The right-side equation is called the relative growth rate. If the population at time $0$ is $P_0$, then the expression for the population is
$$ P(t) = P_0 e^{0.02t} $$Radioactive Decay
Radioactive substances decay by spontaneously emitting radiation. If $m(t)$ is the mass remaining from an initial mass $m_0$ ate time $t$, then the relative decay rate:
$$ -\frac{\frac{\delta m(t)}{\delta t}}{m(t)} $$has experimentally been found to be constant. It follows that
$$ \frac{\frac{\delta m(t)}{\delta t}}{m(t)} = k $$$$ \frac{\delta m(t)}{\delta t} = km(t) $$where $k$ is a negative constant. In other words, radioactive substances decay at a rate proportional to the remaining mass. Therefore
$$ m(t) = m_0 e^{kt} $$Physicists express the rate of decay in terms of half-life, the time required for half of any given quantity to decay.
Newton’s Law of Cooling
Newton’s Law of Cooling states that the rate of cooling of an object is proportional to the temperature difference between the object and its surroundings, if this difference is not too large. If we let $T(t)$ be the temperature of the object at time $t$ and $T_s$ be the temperature of the surroundings, then
$$ \frac{\delta T(t)}{\delta t} = k(T - T_s) $$where $k$ is a constant.
Continuously Compounded Interest
In general, if an amount $A_0$ is invested at an interest rate $r$, then after $t$ years it is worth $A_0(1 + r)^t$. Usually, interest is compounded more frequently, say, $n$ times per year. Then in each compounding perior the rate is $\frac{r}{n}$, and there are $nt$ compounding periods in $t$ years. So the value of the investment is
$$ A = A_0 \left(1 + \frac{r}{n}\right)^{nt} $$If we let $n \to \infty$, then
$$ A(t) = \lim_{n \to \infty} A_0 \left(1 + \frac{r}{n}\right)^{nt} $$$$ = \lim_{n \to \infty} A_0 \left[\left(1 + \frac{r}{n}\right)^{\frac{n}{r}}\right]^{rt} $$$$ = A_0 \left[\lim_{n \to \infty} \left(1 + \frac{r}{n}\right)^{\frac{n}{r}}\right]^{rt} $$If we let $m = \frac{r}{n}$
$$ = A_0 \left[\lim_{m \to \infty} \left(1 + m\right)^{\frac{1}{m}}\right]^{rt} $$But, as we saw before, the limit in this expression is equal to the number $e$ (see Number $e$ as a limit). So witht continuous compounding of interest at interest rate $r$, the amount after $t$ years is
$$ A(t) = A_0e^{rt} $$If we differentiate this equation we get
$$ \frac{\delta A}{\delta t} = rA_0 e^{rt} = rA(t) $$Which says that the rate of increase of an investment is proportional to its size.
Related Rates
If we are pumping air into a balloon, both the volume and the radius of the balloon are increasing and their rates of increase are related to each other. But it is much easier to measure directly the rate of increase of the volume than the rate of increase of the radius.
In a related rates problem the idea is to compute the rate of change of one quantity in terms of the rate of change of another quantity (which may be more easily measured). The procedure is to find an equation that relates the two quantities and then use the Chain Rule to differentiate both sides with respect to time.
Example 1. Air is being pumped into a spherical balloon so that its volume increases at a rate of $100 \text{cm}^3/\text{s}$. How fast is the radius of the balloon increasing when the diameter is $50 \text{cm}$?
Let $V$ be the volume of the balloon and let $r$ be its radius.
$$\frac{\delta V}{\delta t} = 100 \text{cm}^3/\text{s}$$$$\frac{\delta r}{\delta t} \text{ when } r = 25$$We know that
$$V = \frac{4}{3}\pi r^3$$In order to use the given information, we differentiate each side of this equation with respect to $t$.
$$\frac{\delta V}{\delta r} = \frac{\delta V}{\delta r}\frac{\delta r}{\delta t} = 4\pi r^2 \frac{\delta r}{\delta t}$$Now we solve for the unknown quantity
$$\frac{\delta r}{\delta t} = \frac{1}{4 \pi r^2} \frac{\delta V}{\delta t}$$If we let $r = 25$ and $\frac{\delta V}{\delta t} = 100$ we obtain
$$\frac{\delta r}{\delta t} = \frac{1}{4 \pi (25)^2} 100 = \frac{1}{25\pi}$$So the radius of the balloon is increasing at the rate of $\frac{1}{25\pi} \approx 0.0127 \text{cm}/\text{s}$ when the diameter is $50 \text{cm}$.
Linear Approximations and Differentials
A curve lies very close to its tangent line near the point of tangency, which is the basis for a method of finding approximate values of functions.
Linearization and Approximation
We use the tangent line at $(a, f(a))$ as an approximation to the curve $y = f(x)$ when $x$ is near $a$ (See Figure 1).
The linear function whose graph is this tangent line is:
$$ L(x) = f(a) + f'(a)(x - a) $$This is called the linearization of $f$ at $a$. The approximation $f(x) \approx L(x)$ or:
$$ f(x) \approx f(a) + f'(a)(x - a) $$is called the linear approximation or tangent line approximation of $f$ at $a$.
Differentials
If $y = f(x)$, where $f$ is a differentiable function, then the differential $dx$ is an independent variable, that is, $dx$ can be given the value of any real number. The differential $dy$ is then defined in terms of $dx$ by the equation
$$ f'(x) = \frac{dy}{dx} \Leftrightarrow dy = f'(x)dx $$So $dy$ is a dependent variable.
The geometric mieaning of differentials is shown in Figure 5. Let $P(x, f(x))$ and $Q(x + \Delta x, f(x + \Delta x))$ be points on the graph of $f$ and let $dx = \Delta x$. Then the corresponding change in $y$ is
$$ \Delta y = f(x + \Delta x) - f(x) $$The slope of the tangent line $PR$ is the derivative $f’(x)$. Then $dy$ represents the amount that the tangent line rises or falls, that is the change predicted by the tangent line (a linear approximation). Whereas $\Delta y$ represents the amount that the curve $y = f(x)$ rises or falls, the actual change in the function’s value when $x$ changes by an amount $dx$.
In the notation of differentials, the linear approximation $f(x) \approx f(a) + f’(a)(x - a)$ can be re-written by taking $dx = x - a$ so $x = a + dx$
$$ f(a + dx) \approx f(a) + f'(a)dx $$Given $dy = f’(a)dx$
$$ f(a + dx) \approx f(a) + dy $$We can use differentials when estimating errors, for example the error that occurs because of approximate measurements. Although a better picture of the error is given by the relative error, which is computed by dividing the error by the value of the original function:
$$ \frac{\Delta f(x)}{f(x)} \approx \frac{\delta f(x)}{f(x)} $$Hyperbolic Functions
Hyperbolic Functions and Their Derivatives
Certain combinations of the exponential functions $e^x$ and $e^{-x}$ arise frequently in mathematics. They have the same relationship to the hyperbola that the trigonometric functions have to the circle. For this reason they are collectively called hyperbolic functions and individually called hyperbolic sine, hyperbolic cosine, and so on.
Hyperbolic Functions
The graphs of hyperbolic sine and hyperbolic cosine can be sketched using graphical addition as in Figures 1 and 2.
Applications to science and engineering occur whenever an entity such as light, velocity, electricity, or radioactivity is gradually absorbed or extinguished because the decay can be represented by hyperbolic functions. The most famous application is the use of hyperbolic cosine to describe the shape of a hanging wire. It can be proved that if a heavy flexible cable (such as an overhead power line) is suspended between two points at the same height, then it takes the shape of a curve with equation
$$ y = c + a \cosh(\frac{x}{a}) $$called a catenary (see Figure 4).
Another application of hyperbolic functions occurs in the description of ocean waves: the velocity of a water wave with length $L$ moving across a body of water with depth $d$ is modeled by the function
$$ \nu = \sqrt{\frac{gL}{2 \pi} \tanh\left(\frac{2\pi d}{L}\right)} $$where $g$ is the acceleration due gravity (see Figure 5).
Hyperbolic Identities
Let’s reason out the name of “hyperbolic functions”:
If $t$ is any real number, then the point $P(\cos t, \sin t)$ lies on the unit circle $x^2 + y^2 = 1$ because $\cos^2 t + \sin^2 t = 1$. In fact, $t$ can be interpreted as the radian measure of $\angle POQ$ in Figure 6. For this reason the trigonometric functions are sometimes called circular functions.
Likewise, if $t$ is any real number, then the point $P(\cosh t, \sinh t)$ lies on the right branch of the hyperbola $x^2 - y^2 = 1$ because $\cosh^2 t - \sinh^2 t = 1$ and $\cosh t \geq 1$. This time, $t$ does not represent the measure of an angle. However, it turns out that $t$ represents twice the area of the shaded hyperbolic sector in Figure 7, just as in the trigonometric case $t$ represents twice the area of the shaded circular sector in Figure 6.
Derivatives of Hyperbolic Functions
Inverse Hyperbolic Functions and Their Derivatives
You can see from Figures 1 and 3 that $\sinh$ and $\tanh$ are one-to-one functions and so they have inverse functions denoted by $\sinh^{-1}$ and $\tanh^{-1}$. Figure 2 shows that $\cosh$ is not oneto-one, but if we restrict the domain to the interval $[0, \infty]$, then the function $y = \cosh x$ is one-to-one and attains all the values in its range $[1, \infty]$. The inverse hyperbolic cosine function is defined as the inverse of this restricted function.
Inverse Hyperbolic Functions
We sketch the graphs of $\sinh^{-1}$, $\cosh^{-1}$, and $\tanh^{-1}$ in Figures 8, 9, and 10 by referring to Figures 1, 2, and 3
Derivatives of Inverse Hyperbolic Functions