Chapter 6.2 Differentiating Elementary Functions

Derivatives of Inverse Functions

We can use the chain rule to derive the derivative of inverse functions.

For example, \(e^x\) and \(\ln(x)\) are inverse functions.

Recall that if \(f\) and \(f^{-1}\) are inverse functions, then we have that for any \(x\) in the domain of \(f^{-1}\), \[f(f^{-1}(x))=x.\] If we take the derivative of both sides, we will need to use the chain rule to compute the derivative of \(f(f^{-1}(x))\). Doing that gives us \[f'(f^{-1}(x))\cdot (f^{-1}(x))'.\]

Hence \[\begin{align*}\left(f(f^{-1}(x))\right)'&=(x)'\\ f'(f^{-1}(x))\cdot (f^{-1}(x))'&=1\\ (f^{-1}(x))'&=\frac{1}{f'(f^{-1}(x))}.\end{align*}\]

Derivative of the Inverse of a Function

Take \(f\) be a function that is differentiable and invertible and take \(f^{-1}\) be its inverse function. Then, as long as \(f'(f^{-1}(a))\not=0\), \(f^{-1}\) is differentiable at \(a\). In fact, \((f^{-1}(a))'=\frac{1}{f'\left(f^{-1}(a)\right)}\)

Example 1

Show the derivative of \(f(x)=\log_a(x)\) is \[f'(x)=\frac{1}{\ln(a)x}.\] Also show that the derivative of \(f(x)=\ln(x)\) is \[f'(x)=\frac{1}{x}.\]

Take \(f(x)=\exp_a(x)\). Then its derivative is \(f'(x)=\ln(a)\exp_a(x)\) and its inverse is \(f^{-1}(x)=\log_a(x)\). Use the Derivative of the Inverse of Function Theorem:

\[ \begin{align*} \left(\log_a(x)\right)'&=\frac{1}{f'\left(f^{-1}(a)\right)}\\ &=\frac{1}{f'(\log_a(x))}\\ &=\frac{1}{\ln(a)\exp_a(\log_a(x))}\\ &=\frac{1}{\ln(a)x}. \end{align*} \]

In the case of \(\ln(x)=\log_e(x)\), the derivative is \(f'(x)=\tfrac{1}{\ln(e)x}=\tfrac{1}{x}.\)

Example 2

Show the derivative of \(f(x)=\arctan(x)\) is \[f'(x)=\frac{1}{x^2+1}.\]

Take \(f(x)=\tan(x)\). Then its derivative is \(f'(x)=\sec^2(x)\) and its inverse function is \(f^{-1}(x)=\arctan(x)\).

Use the Derivative of the Inverse of Function Theorem: e

\[ \begin{align*} \left( \arctan(a) \right)'&=\frac{1}{f'\left(f^{-1}(a)\right)}\\ &=\frac{1}{f'(\arctan(a))}\\ &=\frac{1}{\sec^2(\arctan(a))}\\ \end{align*} \]

Rewrite this expression by using trigonometry. First, the definition of arctangent states that \(\arctan(x)=y\) if and only if \(x=\tan(y)\) . Construct the following right triangle based on the equation \(x=\tan(y)\):

Use the right triangle to rewrite secant:

\[\sec^2(\arctan(a))=\sec^2(y)=\left(\sec(y)\right)^2=\left(\sqrt{a^2+1}\right)^2=a^2+1.\]

Substitute this to get

\[ \begin{align*} \left( \arctan(a) \right)'&=\frac{1}{\sec^2(\arctan(a))}\\ &=\frac{1}{a^2+1}.\\ \end{align*} \]

Derivative of Inverse Trig Functions

1. Take \(f(x)=\arctan(x)\). Then \((\arctan(x))'=\frac{1}{x^2+1}.\)
2. Take \(f(x)=\arccos(x)\). Then \((\arccos(x))'=-\frac{1}{\sqrt{1-x^2}}.\)
3. Take \(f(x)=\arcsin(x)\). Then \((\arcsin(x))'=\frac{1}{\sqrt{1-x^2}}\)

Example 3

Compute \(f'\) of \(f(x)=\arctan\left(\log_5\left(\tfrac{x+3}{x+1}\right)\right).\)

Take \(f_1(x)=\arctan(x),f_2(x)=\log_5(x),f_3(x)=x+3,f_4(x)=x+1\) so that \[f(x)=f_1\left(f_2\left(\tfrac{f_3(x)}{f_4(x)}\right)\right).\] Take the derivative of each function to get \(f_1'(x)=\tfrac{1}{x^2+1},f_2(x)=\tfrac{1}{\ln(5)x},f_3'(x)=1,\) and \(f_4'(x)=1.\) Use the derivatives rules:

\[ \begin{align*} \left(\arctan\left(\log_5\left(\tfrac{x+3}{x+1}\right)\right)\right)'&=f_1'\left(f_2\left(\tfrac{f_3(x)}{f_4(x)}\right)\right)\cdot \left[f_2\left(\tfrac{f_3(x)}{f_4(x)}\right)\right]'&&\text{chain rule}\\ &=f_1'\left(f_2\left(\tfrac{f_3(x)}{f_4(x)}\right)\right)\cdot f_2'\left(\tfrac{f_3(x)}{f_4(x)}\right)\cdot\left[\tfrac{f_3(x)}{f_4(x)}\right]'&&\text{chain rule}\\ &=f_1'\left(f_2\left(\tfrac{f_3(x)}{f_4(x)}\right)\right)\cdot f_2'\left(\tfrac{f_3(x)}{f_4(x)}\right)\cdot\tfrac{f_3'(x)f_4(x)-f_3(x)f_4'(x)}{(f_4(x))^2}&&\text{quotient rule}\\ &=\frac{1}{\left(f_2\left(\tfrac{f_3(x)}{f_4(x)}\right)\right)^2+1}\cdot \frac{1}{\ln(5)\left[\tfrac{f_3(x)}{f_4(x)}\right]}\cdot\frac{f_3'(x)f_4(x)-f_3(x)f_4'(x)}{(f_4(x))^2}\\ &=\frac{1}{\left(\log_5\left(\tfrac{x+3}{x+1}\right)\right)^2+1}\cdot\frac{1}{\ln(5)\left(\tfrac{x+3}{x+1}\right)}\cdot\frac{1\cdot(x+3)-(x+1)\cdot1}{(x+1)^2}\\ \end{align*} \]

Implicitly Defined Functions and Their Derivatives

So far, we have been working with functions of the form \(y=f(x)\). However, not all curves can be described by functions. In fact, some curves can be described in terms of an equation.

For example, the unit circle centered at the origin can be described by the (implicit) equation \[x^2+y^2=1.\]

If we solve for \(y\), we get \(y=\sqrt{1-x^2}\), which describes the upper half of the circle, and \(y=-\sqrt{1-x^2}\), which describes the lower half of the circle.

So if we wanted to find the tangent line to unit circle at \(\left(\frac{\sqrt{2}}{2} ,\frac{\sqrt{2}}{2}\right)\), we would use \(f(x)=\sqrt{1-x^2}\) as our function, find the derivative of \(f\), evaluate the derivative at \(x=\frac{\sqrt{2}}{2}\) to get the slope of the tangent line, and then use \(L(x)=f(a)+f'(a)(x-a)\) to write the formula for the tangent line.

However, that is not always possible because not every curve that is desribed by an equation can be rewritten into the form \(y=f(x)\). So instead we use implict differentiation.

First some notation. The following all describe the same concept.

• \(f'(x)\)

• \(\tfrac{df}{dx}\) Leibniz notation

• \(Df\) Newton-Leibniz notation

Example 4

Assume that \(y\) is defined implicitly by the equation \(x^2+4y^2=xy+\sin(y)\). Calculate \(\frac{dy}{dx}\).

Implicitly defined means that \(y=y(x)\) is a function of \(x\). So \(\frac{d}{dx}y=\frac{dy(x)}{dx}.\) Because there is no explicit formula for \(y,\) there is no further simplifcation that can be done.

Take the derivative to get:

\[ \begin{align*} x^2+4y^2&=xy+\sin(y)\\ \frac{d}{dx}(x^2+4y^2)&=\frac{d}{dx}(xy+\sin(y))\\ \frac{d}{dx}x^2+\frac{d}{dx}4y^2&=\frac{d}{dx}xy+\frac{d}{dx}\sin(y)\\ 2x+\frac{d}{dx}4(y(x))^2&=\frac{d}{dx}(x\cdot y(x))+\frac{d}{dx}\sin(y(x))&&\text{ remember }y \text{ is a function.} \end{align*} \]

Use the chain rule to get

\[\frac{d}{dx}4(y(x))^2=4\cdot 2\cdot (y(x))^1\cdot \frac{d}{dx}y(x)=8y(x)\frac{dy(x)}{dx}\]

and

\[\frac{d}{dx}\sin(y(x))=\cos(y(x))\cdot \frac{d}{dx}y(x)=\cos(y(x))\frac{dy(x)}{dx}.\]

Use the product rule to get

\[\frac{d}{dx}(x\cdot y(x))=\frac{d}{dx}(x)\cdot y(x)+x\cdot\frac{d}{dx}y(x)=1\cdot y(x)+x\frac{dy(x)}{dx}.\]

Put everything together and simplify:

\[ \begin{align*} 2x+\frac{d}{dx}4(y(x))^2&=\frac{d}{dx}(x\cdot y(x))+\frac{d}{dx}\sin(y(x))\\ 2x+8y\frac{dy}{dx}&=y+x\frac{dy}{dx}+\cos(y)\frac{dy}{dx}. \end{align*} \]

Now solve for \(\tfrac{dy}{dx}\): \[ \begin{align*} 2x+8y\frac{dy}{dx}&=y+x\frac{dy}{dx}+\cos(y)\frac{dy}{dx}\\ 2x-y&=x\frac{dy}{dx}+\cos(y)\frac{dy}{dx}-8y\frac{dy}{dx}\\ 2x-y&=(x+\cos(y)-8y)\frac{dy}{dx}\\ \frac{2x-y}{x+\cos(y)-8y}&=\frac{dy}{dx}. \end{align*} \]

Suppose that the function \(f\) has the property that \(f(x)\) is nonzero.

The chain rule implies that \[ \begin{align*} \frac{{\rm d}}{{\rm d}x}\ln(|f(x)|) = \frac{f^\prime(x)}{f(x)} \quad \text{and so} \quad f^\prime(x) = f(x)\frac{{\rm d}\ln(|f(x)|)}{{\rm d}x}. \end{align*} \]

This allows us to exploit the algebraic properties of the logarithm in order to simplify computing derivatives.

Calculus texts often refer to differentiation performed in this way as logarithmic differentiation.

Logarithm Properties

Let \(b>0\) and \(b\not=1\) and let \(n\) be a real number. Then for \(X>0,Y>0\), we have

1. \(\log_b[XY]=\log_b[X]+\log_b[Y]\)
2. \(\log_b\left[\frac{X}{Y}\right]=\log_b[X]-\log_b[Y]\)
3. \(\log_b[X^n]=n\log_b[X]\)

Example 5

1. \(\ln(x^3)=3\ln(x)\)
2. \(\ln\left(\frac{x}{\tan(x)}\right)=\ln(x)-\ln(\tan(x))\)
3. \(\ln\left(\sqrt{x}\tan(x)\right)=\ln\left(\sqrt{x}\right)+\ln(\tan(x))\)

Example 6

Take \(f(x)=\sqrt[5]{\frac{x^2-9}{x+4}}\). Find \(f'\) by using logarithm differentiation.

Rewrite \(f\) as \(f(x)=\left(\frac{x^2-9}{x+4}\right)^{1/5}\).

Take the logarithm of both sides and rewrite:

\[\begin{align*}f(x)&=\left(\frac{x^2-9}{x+4}\right)^{1/5}\\ \ln(f(x))&=\ln\left(\left(\frac{x^2-9}{x+4}\right)^{1/5}\right)\\ \ln(f(x))&=\frac{1}{5}\ln\left(\frac{x^2-9}{x+4}\right)\\ \ln(f(x))&=\frac{1}{5}\left(\ln\left(x^2-9\right)-\ln\left(x+4\right)\right), \end{align*} \]

Take the derivative of both sides of the equation. All the logarithm terms will involve a chain rule. Take the derivative to get

\[\begin{align*} \frac{1}{f(x)}f'(x)&=\frac{1}{5}\left(\frac{1}{x^2-9}\cdot 2x-\frac{1}{x+4}\cdot 1\right). \end{align*}\]

Multiply both sides by \(f(x)\) and plugging in \(f(x)=\left(\frac{x^2-9}{x+4}\right)^{1/5}\) to get \[\begin{align*} f'(x)&=\frac{1}{5}\left(\frac{x^2-9}{x+4}\right)^{1/5}\left(\frac{2x}{x^2-9}-\frac{1}{x+4}\right). \end{align*}\]

For any positive differentiable function \(f\), rather than use logarithmic differentiation, it can be simpler to use the formula \[f(x) = e^{\ln(f(x))}\] in order to transform the differentiation problem into something simpler.

For instance, take \(f\) to be the function that is defined for all \(x\) in \((0, \infty)\) by \[f(x) = x^x.\] What is \(f^\prime(x)\)?

Differentiate both sides of the above equality to see that \[\begin{align*} \frac{{\rm d}}{{\rm d}x} x^x &= \frac{{\rm d}}{{\rm d}x} e^{\ln(x^x)} \\&= \frac{{\rm d}}{{\rm d}x} e^{x\ln(x)} \\&= e^{x\ln(x)} (x\ln(x))' \\& = e^{x\ln(x)}\left(\ln(x) +1\right) = (\ln(x) +1)x^x.\end{align*}\]

Example 7

For each function \(f\) that is given below, determine \(f^\prime(x)\).

1. \(f(x)=x^{x^2+5}\)
2. \(f(x)=(\tan(x))^x\)

1. Rewrite using the formula \(f(x) = e^{\ln(f(x))}\) and take the derivative to get

\[\begin{align*} \frac{{\rm d}}{{\rm d}x} x^{x^2+5} &= \frac{{\rm d}}{{\rm d}x} e^{\ln(x^{x^2+5})} \\&= \frac{{\rm d}}{{\rm d}x} e^{(x^2+5)\ln(x)} \\&= e^{(x^2+5)\ln(x)}\cdot((x^2+5)\ln(x))'\\ &=e^{(x^2+5)\ln(x)}\cdot\left(2x\ln(x)+\frac{x^2+5}{x}\right)\\ &=e^{\ln(x^{x^2+5})}\cdot\left(2x\ln(x)+\frac{x^2+5}{x}\right)\\ &=x^{x^2+5}\cdot\left(2x\ln(x)+\frac{x^2+5}{x}\right)\\ \end{align*}\]

2. Rewrite using the formula \(f(x) = e^{\ln(f(x))}\) and take the derivative to get

\[\begin{align*} \frac{{\rm d}}{{\rm d}x} (\tan(x))^x &= \frac{{\rm d}}{{\rm d}x} e^{\ln((\tan(x))^x)} \\&= \frac{{\rm d}}{{\rm d}x} e^{(x)\ln(\tan(x))} \\&= e^{x\ln(\tan(x))}\cdot(x\ln(\tan(x)))'\\ &=e^{x\ln(\tan(x))}\cdot\left(\ln(\tan(x))+x\cdot\frac{\sec^2(x)}{\tan(x)}\right)\\ &=e^{\ln(\tan(x))^x}\cdot\left(\ln(\tan(x))+x\cdot\frac{\sec^2(x)}{\tan(x)}\right)\\ &=(\tan(x))^x\cdot\left(\ln(\tan(x))+x\cdot\frac{\sec^2(x)}{\tan(x)}\right)\\ \end{align*}\]

Related Rates Problems

Many problems that appear in applications involve covarying quantities \(x\) and \(y\), this is to say that the set of pairs \((x,y)\) is the level set of a function of two variables.

In this case, one of the variables \(x\) and \(y\) might only be implicitly given as a function of the other variable.

Of course, one variable may indeed be explicitly given as a functions of the other.

In either case, implicit differentiation is an important tool for determining the local linear approximation of the physical quantities.

A spherical balloon is being inflated with air at a rate so its volume is increasing at a rate of \(100\) centimeters cubed per second. Determine the rate of change of the radius of the balloon when the diameter is \(50\) centimeters.

Because the shape of the balloon is spherical, use the volume formula of a sphere to determine the volume of the balloon. To find how fast the radius of the balloon increases, find the derivative of the radius. Define the following:

Take \(V=V(t)\) to be the volume of the balloon at time \(t\). Take \(r=r(t)\) be the radius of the balloon at time \(t\).

Relate the functions in the following way: the volume of a sphere is given by \(V=\frac{4}{3}\pi r^3\). This is an equation that relates \(V\) and \(r\): \(V(t)=\frac{4}{3}\pi (r(t))^3\). To find the derivative of \(r\), take the derivative of both sides with respect to \(t\) :

\[ \begin{align*} (V(t))'&=(\frac{4}{3}\pi (r(t))^3)'\\ V'(t)&=\frac{4}{3}\pi ((r(t))^3)'\\ V'(t)&=\frac{4}{3}\pi 3(r(t))^2(r(t))'\\ V'(t)&=4\pi (r(t))^2r'(t).\\ \end{align*} \]

(use the chain rule to compute the derivative of \(r^3=(r(t))^3\))

Solve for \(r'\) to get \(r'(t)=\frac{V'(t)}{4\pi (r(t))^2}\). The goal is now to find how fast the radius of the balloon increases when the diameter is 50 cm and \(V'=100\) (volume increases at a rate of 100 centimeters cubed per second). This means that at that particular time, \(r=25\) centimeters (the radius is half the diameter). Plug that into our formula for \(r'\) to get \(r'(t)=\frac{100}{4\pi (25)^2}=\frac{1}{25\pi}\approx 0.013\) centimeters per second.

There are two cars, Car A and Car B, that are headed towards the same intersection. Car A travels west at \(50\) miles per hour and Car B travels north at \(60\) miles per hour. Determine the rate at which the cars approach each other when Car A is \(0.3\) miles from the intersection and Car B is \(0.4\) miles from the intersection.

Take \(x(t)\) be the distance from Car A to the intersection, \(y(t)\) be the distance from Car B to the intersection, and \(z(t)\) be the distance from Car A to Car B.

Find \(z'(t)\), which is the rate the cars are approaching each other. Use Pythagorean theorem to relate the quantities: \[x^2(t)+y^2(t)=z^2(t)\]

Take the derivative of both sides: \[\begin{align*}(x^2(t)+y^2(t))'&=(z^2(t))'\\ 2x(t)x'(t)+2y(t)'y(t)&=2z(t)z'(t)\end{align*}\]

Solve for \(z'\) to get \(z'(t)=\frac{1}{2z(t)}\left(2x(t)x'(t)+2y(t)y'(t)\right)\) .

The problem states that at a particular time, \(x'(t)=-50\) mph, \(y'(t)=-60\) mph, \(x(t)=0.3\) m, \(y(t)=0.4\)m. The rates are negative because the distance between Car A and intersection is decreasing and the distance between Car B and intersection is decreasing. The value of \(z(t)\) is unknown. However, use Pythagorean theorem to solve for \(z\).

That is solve for \(z\) in the equation \((0.3)^2+(0.4)^2=z^2(t)\) to get \(z(t)=0.5\) m. Plug in all relevant values to get that

\[z'(t)=\frac{1}{2(0.5)}\left(2\cdot(0.3)\cdot(-50)+2\cdot(0.4)\cdot(-60)\right)\approx -78\text{mph}\]

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