The Derivative of an Inverse Function

We begin by considering a function and its inverse. If $f(x)$ is both invertible and differentiable, it seems reasonable that the inverse of $f(x)$ is also differentiable. (Figure) shows the relationship between a function $f(x)$ and its inverse $f^{-1}(x)$. Look at the point $(a,f^{-1}(a))$ on the graph of $f^{-1}(x)$ having a tangent line with a slope of $(f^{-1})^{\prime}(a)=\frac{p}{q}$. This point corresponds to a point $(f^{-1}(a),a)$ on the graph of $f(x)$ having a tangent line with a slope of $f^{\prime}(f^{-1}(a))=\frac{q}{p}$. Thus, if $f^{-1}(x)$ is differentiable at $a$, then it must be the case that

Figure 1. The tangent lines of a function and its inverse are related; so, too, are the derivatives of these functions.

We may also derive the formula for the derivative of the inverse by first recalling that $x=f(f^{-1}(x))$. Then by differentiating both sides of this equation (using the chain rule on the right), we obtain

Solving for $(f^{-1})^{\prime}(x)$, we obtain

We summarize this result in the following theorem.

From the previous example, we see that we can use the inverse function theorem to extend the power rule to exponents of the form $\frac{1}{n}$, where $n$ is a positive integer. This extension will ultimately allow us to differentiate $x^q$, where $q$ is any rational number.

Derivative of the Exponential Function

Just as when we found the derivatives of other functions, we can find the derivatives of exponential and logarithmic functions using formulas. As we develop these formulas, we need to make certain basic assumptions. The proofs that these assumptions hold are beyond the scope of this course.

First of all, we begin with the assumption that the function $B(x)=b^x, \, b>0$, is defined for every real number and is continuous. In previous courses, the values of exponential functions for all rational numbers were defined—beginning with the definition of $b^n$, where $n$ is a positive integer—as the product of $b$ multiplied by itself $n$ times. Later, we defined $b^0=1, \, b^{−n}=\frac{1}{b^n}$ for a positive integer $n$, and $b^{s/t}=(\sqrt[t]{b})^s$ for positive integers $s$ and $t$. These definitions leave open the question of the value of $b^r$ where $r$ is an arbitrary real number. By assuming the continuity of $B(x)=b^x, \, b>0$, we may interpret $b^r$ as $\underset{x\to r}{\lim}b^x$ where the values of $x$ as we take the limit are rational. For example, we may view ${4}^{\pi}$ as the number satisfying

As we see in the following table, $4^{\pi}\approx 77.88$.

We also assume that for $B(x)=b^x, \, b>0$, the value $B^{\prime}(0)$ of the derivative exists. In this section, we show that by making this one additional assumption, it is possible to prove that the function $B(x)$ is differentiable everywhere.

We make one final assumption: that there is a unique value of $b>0$ for which $B^{\prime}(0)=1$. We define $e$ to be this unique value, as we did in Introduction to Functions and Graphs. (Figure) provides graphs of the functions $y=2^x, \, y=3^x, \, y=2.7^x$, and $y=2.8^x$. A visual estimate of the slopes of the tangent lines to these functions at 0 provides evidence that the value of $e$ lies somewhere between 2.7 and 2.8. The function $E(x)=e^x$ is called the natural exponential function. Its inverse, $L(x)=\log_e x=\ln x$ is called the natural logarithmic function.

Figure 1. The graph of $E(x)=e^x$ is between $y=2^x$ and $y=3^x$.

For a better estimate of $e$, we may construct a table of estimates of $B^{\prime}(0)$ for functions of the form $B(x)=b^x$. Before doing this, recall that

for values of $x$ very close to zero. For our estimates, we choose $x=0.00001$ and $x=-0.00001$ to obtain the estimate

See the following table.

The evidence from the table suggests that $2.7182<e<2.7183$.

The graph of $E(x)=e^x$ together with the line $y=x+1$ are shown in (Figure). This line is tangent to the graph of $E(x)=e^x$ at $x=0$.

Figure 2. The tangent line to $E(x)=e^x$ at $x=0$ has slope 1.

Now that we have laid out our basic assumptions, we begin our investigation by exploring the derivative of $B(x)=b^x, \, b>0$. Recall that we have assumed that $B^{\prime}(0)$ exists. By applying the limit definition to the derivative we conclude that

Turning to $B^{\prime}(x)$, we obtain the following.

We see that on the basis of the assumption that $B(x)=b^x$ is differentiable at $0, \, B(x)$ is not only differentiable everywhere, but its derivative is

For $E(x)=e^x, \, E^{\prime}(0)=1$. Thus, we have $E^{\prime}(x)=e^x$. (The value of $B^{\prime}(0)$ for an arbitrary function of the form $B(x)=b^x, \, b>0$, will be derived later.)

Derivative of the Logarithmic Function

Now that we have the derivative of the natural exponential function, we can use implicit differentiation to find the derivative of its inverse, the natural logarithmic function.

Proof

If $x>0$ and $y=\ln x$, then $e^y=x$. Differentiating both sides of this equation results in the equation

$e^y\frac{dy}{dx}=1$.

Solving for $\frac{dy}{dx}$ yields

$\frac{dy}{dx}=\frac{1}{e^y}$.

Finally, we substitute $x=e^y$ to obtain

$\frac{dy}{dx}=\frac{1}{x}$.

We may also derive this result by applying the inverse function theorem, as follows. Since $y=g(x)=\ln x$ is the inverse of $f(x)=e^x$, by applying the inverse function theorem we have

$\frac{dy}{dx}=\frac{1}{f^{\prime}(g(x))}=\frac{1}{e^{\ln x}}=\frac{1}{x}$.

Using this result and applying the chain rule to $h(x)=\ln(g(x))$ yields

$h^{\prime}(x)=\frac{1}{g(x)} g^{\prime}(x). _\blacksquare$

The graph of $y=\ln x$ and its derivative $\frac{dy}{dx}=\frac{1}{x}$ are shown in (Figure).

Figure 3. The function $y=\ln x$ is increasing on $(0,+\infty)$. Its derivative $y^{\prime} =\frac{1}{x}$ is greater than zero on $(0,+\infty)$.

Taking a Derivative of a Natural Logarithm

Find the derivative of $f(x)=\ln(x^3+3x-4)$.

Answer:

Use (Figure) directly.

$\begin{array}{lllll} f^{\prime}(x) & =\frac{1}{x^3+3x-4} \cdot (3x^2+3) & & & \text{Use} \, g(x)=x^3+3x-4 \, \text{in} \, h^{\prime}(x)=\frac{1}{g(x)} g^{\prime}(x). \\ & =\frac{3x^2+3}{x^3+3x-4} & & & \text{Rewrite.} \end{array}$

Using Properties of Logarithms in a Derivative

Find the derivative of $f(x)=\ln(\frac{x^2 \sin x}{2x+1})$.

Answer:

At first glance, taking this derivative appears rather complicated. However, by using the properties of logarithms prior to finding the derivative, we can make the problem much simpler.

$\begin{array}{lllll} f(x) & = \ln(\frac{x^2 \sin x}{2x+1})=2\ln x+\ln(\sin x)-\ln(2x+1) & & & \text{Apply properties of logarithms.} \\ f^{\prime}(x) & = \frac{2}{x} + \frac{\cos x}{\sin x} -\frac{2}{2x+1} & & & \text{Apply sum rule and} \, h^{\prime}(x)=\frac{1}{g(x)} g^{\prime}(x). \\ & = \frac{2}{x} + \cot x - \frac{2}{2x+1} & & & \text{Simplify using the quotient identity for cotangent.} \end{array}$

Differentiate: $f(x)=\ln (3x+2)^5$.

Answer:

$f^{\prime}(x)=\frac{15}{3x+2}$

Hint

Use a property of logarithms to simplify before taking the derivative.

Now that we can differentiate the natural logarithmic function, we can use this result to find the derivatives of $y=\log_b x$ and $y=b^x$ for $b>0, \, b\ne 1$.

Derivatives of General Exponential and Logarithmic Functions

Let $b>0, \, b\ne 1$, and let $g(x)$ be a differentiable function.

1. If $y=\log_b x$, then
   $\frac{dy}{dx}=\frac{1}{x \ln b}$.
   More generally, if $h(x)=\log_b (g(x))$, then for all values of $x$ for which $g(x)>0$,
   $h^{\prime}(x)=\frac{g^{\prime}(x)}{g(x) \ln b}$.
2. If $y=b^x$, then
   $\frac{dy}{dx}=b^x \ln b$.
   More generally, if $h(x)=b^{g(x)}$, then
   $h^{\prime}(x)=b^{g(x)} g^{\prime}(x) \ln b$.

Logarithmic Differentiation

At this point, we can take derivatives of functions of the form $y=(g(x))^n$ for certain values of $n$, as well as functions of the form $y=b^{g(x)}$, where $b>0$ and $b\ne 1$. Unfortunately, we still do not know the derivatives of functions such as $y=x^x$ or $y=x^{\pi}$. These functions require a technique called logarithmic differentiation, which allows us to differentiate any function of the form $h(x)=g(x)^{f(x)}$. It can also be used to convert a very complex differentiation problem into a simpler one, such as finding the derivative of $y=\frac{x\sqrt{2x+1}}{e^x \sin^3 x}$. We outline this technique in the following problem-solving strategy.