3.9 Derivatives of Exponential and Logarithmic Functions

So far, we have learned how to differentiate a variety of functions, including trigonometric, inverse, and implicit functions. In this section, we explore derivatives of exponential and logarithmic functions. As we discussed in Introduction to Functions and Graphs, exponential functions play an important role in modeling population growth and the decay of radioactive materials. Logarithmic functions can help rescale large quantities and are particularly helpful for rewriting complicated expressions.

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 $b^{s / t} = (t b)^{s}$ 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 $lim_{x \to r} b^{x}$ where the values of $x$ as we take the limit are rational. For example, we may view $4^{π}$ as the number satisfying

4^{3} < 4^{π} < 4^{4}, 4^{3.1} < 4^{π} < 4^{3.2}, 4^{3.14} < 4^{π} < 4^{3.15}, 4^{3.141} < 4^{π} < 4^{3.142}, 4^{3.1415} < 4^{π} < 4^{3.1416}, \dots .

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

Table 3.6Approximating a Value of

4^{π}

$x$	$4^{x}$	$x$	$4^{x}$
$3$

We also assume that for $B (x) = b^{x}, b > 0,$ the value $B^{'} (0)$ 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 $B^{'} (0) = 1.$ We define $e$ to be this unique value, as we did in Introduction to Functions and Graphs. Figure 3.33 provides graphs of the functions $y = 2^{x}, y = 3^{x}, y = 2.7^{x},$ and A visual estimate of the slopes of the tangent lines to these functions at 0 provides evidence that the value of lies somewhere between 2.7 and 2.8. The function is called the . Its inverse, is called the .

The graphs of 3x, 2.8x, 2.7x, and 2x are shown. In quadrant I, their order from least to greatest is 2x, 2.7x, 2.8x, and 3x. In quadrant II, this order is reversed. All cross the y-axis at (0, 1). — Figure 3.33The 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 $B^{'} (0)$ for functions of the form $B (x) = b^{x} .$ Before doing this, recall that

B^{'} (0) = x \to 0 lim \frac{b ^{x} - b ^{0}}{x - 0} = x \to 0 lim \frac{b ^{x} - 1}{x} \approx \frac{b ^{x} - 1}{x}

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

\frac{b ^{- 0.00001} - 1}{- 0.00001} < B^{'} (0) < \frac{b ^{0.00001} - 1}{0.00001} .

See the following table.

Table 3.7Estimating a Value of

e

$b$	$\frac{b ^{- 0.00001} - 1}{- 0.00001} < B^{'} (0) < \frac{b ^{0.00001} - 1}{0.00001}$

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 3.34. This line is tangent to the graph of $E (x) = e^{x}$ at $x = 0.$

Graph of the function ex along with its tangent at (0, 1), x + 1. — Figure 3.34The 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 $B^{'} (0)$ exists. By applying the limit definition to the derivative we conclude that

B^{'} (0) = h \to 0 lim \frac{b ^{0 + h} - b ^{0}}{h} = h \to 0 lim \frac{b ^{h} - 1}{h} .

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

B^{'} (x) = lim_{h \to 0} \frac{b ^{x + h} - b ^{x}}{h} = lim_{h \to 0} \frac{b ^{x} b ^{h} - b ^{x}}{h} = lim_{h \to 0} \frac{b ^{x} ( b ^{h} - 1 )}{h} = b^{x} lim_{h \to 0} \frac{b ^{h} - 1}{h} = b^{x} B^{'} (0) Apply the limit definition of the derivative. Note that b^{x + h} = b^{x} b^{h} . Factor out b^{x} . Apply a property of limits. Use B^{'} (0) = lim_{h \to 0} \frac{b ^{0 + h} - b ^{0}}{h} = lim_{h \to 0} \frac{b ^{h} - 1}{h} .

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

B^{'} (x) = b^{x} B^{'} (0) .

For $E (x) = e^{x}, E^{'} (0) = 1.$ Thus, we have $E^{'} (x) = e^{x} .$ (The value of for an arbitrary function of the form 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{d y}{d x} = 1.

Solving for $\frac{d y}{d x}$ yields

\frac{d y}{d x} = \frac{1}{e ^{y}} .

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

\frac{d y}{d x} = \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{d y}{d x} = \frac{1}{f ^{'} ( g ( x ))} = \frac{1}{e ^{l n x}} = \frac{1}{x} .

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

h^{'} (x) = \frac{1}{g ( x )} g^{'} (x) .

□

The graph of $y = ln x$ and its derivative $\frac{d y}{d x} = \frac{1}{x}$ are shown in Figure 3.35.

Graph of the function ln x along with its derivative 1/x. The function ln x is increasing on (0, + ∞). Its derivative is decreasing but greater than 0 on (0, + ∞). — Figure 3.35 $The function y = ln x$ is increasing on $(0, + \infty) .$ Its derivative $y^{'} = \frac{1}{x}$ is greater than zero on

Now that we can differentiate the natural logarithmic function, we can use this result to find the derivatives of $y = l o g_{b} x$ and $y = b^{x}$ for $b > 0, b \neq = 1.$

Proof

If $y = lo g_{b} x,$ then $b^{y} = x .$ It follows that $ln (b^{y}) = ln x .$ Thus Solving for we have Differentiating and keeping in mind that is a constant, we see that

\frac{d y}{d x} = \frac{1}{x ln b} .

The derivative in Derivatives of Exponential and Logarithmic Functions now follows from the chain rule.

If $y = b^{x},$ then $ln y = x ln b .$ Using implicit differentiation, again keeping in mind that $ln b$ is constant, it follows that $\frac{1}{y} \frac{d y}{d x} = ln b .$ Solving for and substituting we see that

\frac{d y}{d x} = y ln b = b^{x} ln b .

The more general derivative (Derivatives of Exponential and Logarithmic Functions) follows from the chain rule.

□

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 Unfortunately, we still do not know the derivatives of functions such as or These functions require a technique called , which allows us to differentiate any function of the form It can also be used to convert a very complex differentiation problem into a simpler one, such as finding the derivative of We outline this technique in the following problem-solving strategy.

Key Concepts

On the basis of the assumption that the exponential function $y = b^{x}, b > 0$ is continuous everywhere and differentiable at 0, this function is differentiable everywhere and there is a formula for its derivative.
We can use a formula to find the derivative of $y = ln x,$ and the relationship $lo g_{b} x = \frac{l n x}{l n b}$ allows us to extend our differentiation formulas to include logarithms with arbitrary bases.

Key Equations

Derivative of the natural exponential function

\frac{d}{d x} (e^{g (x)}) = e^{g (x)} g^{'} (x)

Derivative of the Exponential Function

4^{3} < 4^{π} < 4^{4}, 4^{3.1} < 4^{π} < 4^{3.2}, 4^{3.14} < 4^{π} < 4^{3.15}, 4^{3.141} < 4^{π} < 4^{3.142}, 4^{3.1415} < 4^{π} < 4^{3.1416}, \dots .

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

Table 3.6Approximating a Value of

4^{π}

$x$	$4^{x}$	$x$	$4^{x}$
$3$

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

B^{'} (0) = x \to 0 lim \frac{b ^{x} - b ^{0}}{x - 0} = x \to 0 lim \frac{b ^{x} - 1}{x} \approx \frac{b ^{x} - 1}{x}

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

\frac{b ^{- 0.00001} - 1}{- 0.00001} < B^{'} (0) < \frac{b ^{0.00001} - 1}{0.00001} .

See the following table.

Table 3.7Estimating a Value of

e

$b$	$\frac{b ^{- 0.00001} - 1}{- 0.00001} < B^{'} (0) < \frac{b ^{0.00001} - 1}{0.00001}$

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 3.34. This line is tangent to the graph of $E (x) = e^{x}$ at $x = 0.$

B^{'} (0) = h \to 0 lim \frac{b ^{0 + h} - b ^{0}}{h} = h \to 0 lim \frac{b ^{h} - 1}{h} .

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

B^{'} (x) = lim_{h \to 0} \frac{b ^{x + h} - b ^{x}}{h} = lim_{h \to 0} \frac{b ^{x} b ^{h} - b ^{x}}{h} = lim_{h \to 0} \frac{b ^{x} ( b ^{h} - 1 )}{h} = b^{x} lim_{h \to 0} \frac{b ^{h} - 1}{h} = b^{x} B^{'} (0) Apply the limit definition of the derivative. Note that b^{x + h} = b^{x} b^{h} . Factor out b^{x} . Apply a property of limits. Use B^{'} (0) = lim_{h \to 0} \frac{b ^{0 + h} - b ^{0}}{h} = lim_{h \to 0} \frac{b ^{h} - 1}{h} .

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

B^{'} (x) = b^{x} B^{'} (0) .

For $E (x) = e^{x}, E^{'} (0) = 1.$ Thus, we have $E^{'} (x) = e^{x} .$ (The value of for an arbitrary function of the form 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{d y}{d x} = 1.

Solving for $\frac{d y}{d x}$ yields

\frac{d y}{d x} = \frac{1}{e ^{y}} .

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

\frac{d y}{d x} = \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{d y}{d x} = \frac{1}{f ^{'} ( g ( x ))} = \frac{1}{e ^{l n x}} = \frac{1}{x} .

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

h^{'} (x) = \frac{1}{g ( x )} g^{'} (x) .

□

The graph of $y = ln x$ and its derivative $\frac{d y}{d x} = \frac{1}{x}$ are shown in Figure 3.35.

Now that we can differentiate the natural logarithmic function, we can use this result to find the derivatives of $y = l o g_{b} x$ and $y = b^{x}$ for $b > 0, b \neq = 1.$

Proof

If $y = lo g_{b} x,$ then $b^{y} = x .$ It follows that $ln (b^{y}) = ln x .$ Thus Solving for we have Differentiating and keeping in mind that is a constant, we see that

\frac{d y}{d x} = \frac{1}{x ln b} .

The derivative in Derivatives of Exponential and Logarithmic Functions now follows from the chain rule.

\frac{d y}{d x} = y ln b = b^{x} ln b .

The more general derivative (Derivatives of Exponential and Logarithmic Functions) follows from the chain rule.

□

Logarithmic Differentiation

Key Concepts

On the basis of the assumption that the exponential function $y = b^{x}, b > 0$ is continuous everywhere and differentiable at 0, this function is differentiable everywhere and there is a formula for its derivative.
We can use a formula to find the derivative of $y = ln x,$ and the relationship $lo g_{b} x = \frac{l n x}{l n b}$ allows us to extend our differentiation formulas to include logarithms with arbitrary bases.

Key Equations

Derivative of the natural exponential function

\frac{d}{d x} (e^{g (x)}) = e^{g (x)} g^{'} (x)