Special derivatives | Derivative basics | Achievable AP Calculus AB

What you’ll learn:

Derivatives of exponential, logarithmic, and trigonometry functions

So far, we’ve worked with the derivatives of polynomials and rational functions. But there are a wide variety of other functions that exist. In this section, we’ll find special derivatives - those of exponential, logarithmic, and trigonometric functions.

Exponential functions

Exponential functions model rapid growth and decay and commonly appear in population models, radioactive decay, and situations with compounded interest. The most important exponential function is the natural exponential function $f (x) = e^{x}$ , which has a key property - its rate of change is equal to itself.

$\frac{d}{d x} e^{x} = e^{x}$

More generally, for any basic exponential function of the form $f (x) = b^{x}$ (where $b > 0$ and $\neq = 1$ ), its derivative is:

$\frac{d}{d x} (b^{x}) = b^{x} \cdot ln (b)$

In other words, just rewrite the function and multiply by the natural log of the base value.

Example

1. Determine the slope of the tangent line to $f (x) = 7^{x}$ at $x = 0$ .

Solution

(spoiler)

Think derivative with the keywords “slope of the tangent line.”

$f^{'} (x) = 7^{x} \cdot ln (7)$

At $x = 0$ , the slope of the tangent line is $f^{'} (0) = 7^{0} ln (7) = ln (7)$ .

2. Where on the graph of

$f (x) = x^{2} e^{x}$

will there be horizontal tangent lines?

Solution

(spoiler)

Answer: At $(0, 0)$ and $(- 2, 4 e^{- 2})$

Using the product rule,

$f^{'} (x) = x^{2} (e^{x}) + e^{x} (2 x)$

A horizontal tangent line means the derivative $= 0$ at that point. Solve for:

$x^{2} e^{x} + 2 x e^{x} = 0$

Make sure to avoid simply moving one of the terms to the other side and dividing both sides by $x e^{x}$ , since that may eliminate a solution. Instead, always factor:

$x e^{x} (x + 2) = 0$

The graph of $y = e^{x}$ is always above the $x$ -axis and there is no solution for which $e^{x} = 0$ . So the two solutions are

$x = 0, - 2$

with corresponding $y$ -coordinates of

$f (0) = (0)^{2} \cdot e^{0} = 0$

$f (- 2) = (- 2)^{2} \cdot e^{- 2} = 4 e^{- 2} \approx 0.544$

Then the two points on $f (x)$ that have a horizontal tangent line are

$(0, 0) and (- 2, 4 e^{- 2})$

Sidenote

Power rule vs. derivatives of exponentials

Be careful not to confuse these two! Given $x$ is a variable and $b$ is a constant:

Power rule if $f (x) = x^{b}$ :
- e.g. $\frac{d}{d x} x^{3} = 3 x^{2}$
Rule for exponentials if $f (x) = b^{x}$
- e.g. $\frac{d}{d x} 3^{x} = 3^{x} ln (3)$

Logarithmic functions

For $f (x) = ln (x)$ :

$\frac{d}{d x} ln (x) = \frac{1}{x}$

Any logarithmic function $f (x) = lo g_{b} x$ (where the base $b > 0$ and $\neq = 1$ ) can be rewritten using the change-of-base rule into

$f (x) = \frac{ln ( x )}{ln ( b )}$

$= \frac{1}{ln ( b )} \cdot ln (x)$

$\frac{1}{ln ( b )}$ is pulled to. the front because it’s a constant. Then the derivative is

$\frac{d}{d x} lo g_{b} x = \frac{1}{x \cdot ln ( b )}$

Notice that $lo g_{e} (x) = ln (x)$ and the derivative is $\frac{1}{x \cdot ln ( e )} = \frac{1}{x}$ .

Sidenote

On the derivative's domain

The derivative is only valid in the domain of the original function. For all logarithmic functions, the argument must be $> 0$ .

Examples

2. Let $f (x) = lo g_{1/2} (x)$ . Determine for which values of $x$ the slope of the tangent line to $f (x)$ is positive and the values for which it is negative.

(spoiler)

Answer: Negative slope for all $x$ -values in $(0, \infty)$ .

$f^{'} (x) = \frac{1}{x \cdot ln ( 1/2 )}$

$= \frac{1}{ln ( 1/2 )} \cdot \frac{1}{x}$

$\approx - 1.44 \cdot \frac{1}{x}$

Because the domain of $f (x)$ is $x > 0$ , so is the domain of $f^{'} (x)$ . No matter what (positive) value of $x$ is inputted, the derivative, or the slope of the tangent line to $f (x)$ , is negative.

Graphing $lo g_{1/2} (x)$ confirms the function is always decreasing as we move from left to right.

3. Find the derivative of

$f (x) = lo g_{x} (4)$

Solution

(spoiler)

Using the change of base formula,

$f (x) = \frac{ln ( 4 )}{ln ( x )}$

Differentiating with the quotient rule,

$f^{'} (x) = \frac{ln ( x ) ( 0 ) - ln ( 4 ) \cdot \frac{1}{x}}{( ln ( x ) ) ^{2}}$

$= - \frac{ln ( 4 )}{x ( ln ( x ) ) ^{2}}$

Here’s a way to confirm that this is the derivative in Desmos:

1. Type

$\frac{d}{d x} (the original function)$

2. Type in the derivative function we found. Use as many parentheses as possible - it’s better to err on the side of too many than not enough.

3. Check that the graphs match.

Trigonometric functions

Let’s start with the basic trig functions: $sin (x)$ and $cos (x)$ .

$\frac{d}{d x} sin (x) = cos (x)$

$\frac{d}{d x} cos (x) = - sin (x)$

From there, the derivative of $tan (x)$ can be derived using the quotient rule (or product rule, if so inclined) by rewriting $tan (x)$ as $\frac{sin ( x )}{cos ( x )}$ .

1. Find the derivative of $f (x) = tan (x)$ by first rewriting it as a quotient.

Solution

(spoiler)

Answer: $sec^{2} (x)$

Using the quotient rule on $f (x) = \frac{sin ( x )}{cos ( x )}$ ,

$f^{'} (x) = \frac{cos ( x ) \cdot cos ( x ) - sin ( x ) \cdot ( - sin ( x ))}{cos ^{2} ( x )}$

$= \frac{cos ^{2} ( x ) + sin ^{2} ( x )}{cos ^{2} ( x )}$

Using the Pythagorean identity
$sin^{2} (x) + cos^{2} (x) = 1$ ,

$f^{'} (x) = \frac{1}{cos ^{2} ( x )} = sec^{2} (x)$

Let’s do the same for the derivative of $cot (x)$ .

2. Find the derivative of $f (x) = cot (x)$ by rewriting it as a quotient.

Solution

(spoiler)

Answer: $- csc^{2} (x)$

Using the quotient rule on $f (x) = \frac{cos ( x )}{sin ( x )}$ ,

$f^{'} (x) = \frac{sin ( x ) \cdot ( - sin ( x )) - cos ( x ) \cdot cos ( x )}{sin ^{2} ( x )}$

$= \frac{- sin ^{2} ( x ) - cos ^{2} ( x )}{sin ^{2} ( x )}$

$= \frac{- ( sin ^{2} ( x ) + cos ^{2} ( x ))}{sin ^{2} ( x )}$

$= \frac{- 1}{sin ^{2} ( x )} = - csc^{2} (x)$

Let’s do the same for the remaining two reciprocal functions:

3. Find the derivatives of

a. $a (x) = sec (x)$
b. $b (x) = csc (x)$

Solutions

(spoiler)

a. Since $sec (x) = \frac{1}{cos ( x )},$

$a^{'} (x) = \frac{cos ( x ) ( 0 ) - 1 ( - sin ( x ))}{cos ^{2} ( x )}$

$= \frac{sin ( x )}{cos ^{2} ( x )}$

$= \frac{1}{cos ( x )} \cdot \frac{sin ( x )}{cos ( x )}$

$= sec (x) tan (x)$

b. Since $csc (x) = \frac{1}{sin ( x )},$

$a^{'} (x) = \frac{sin ( x ) ( 0 ) - 1 ( cos ( x ))}{sin ^{2} ( x )}$

$= \frac{- cos ( x )}{sin ^{2} ( x )}$

$= \frac{- 1}{sin ( x )} \cdot \frac{cos ( x )}{sin ( x )}$

$= - csc (x) cot (x)$

Memorizing the trig derivatives will save time but it’s also handy to know how to derive them from the derivatives of $sin (x)$ and $cos (x)$ .

Key points

These tables summarize the special derivatives:

For exponential and logarithmic functions:

$f (x)$	$f^{'} (x)$
$e^{x}$	$e^{x}$
$a^{x}$	$a^{x} ln (a)$
$ln (x)$	$\frac{1}{x}$
$lo g_{b} (x)$	$\frac{1}{x ln ( b )}$

For trig functions:

$f (x)$	$f^{'} (x)$
$sin (x)$	$cos (x)$
$tan (x)$	$sec^{2} (x)$
$cot (x)$	$- csc^{2} (x)$
$sec (x)$	$sec (x) tan (x)$
$csc (x)$	$- csc (x) cot (x)$

What you’ll learn:

Derivatives of exponential, logarithmic, and trigonometry functions

Exponential functions

$\frac{d}{d x} e^{x} = e^{x}$

More generally, for any basic exponential function of the form $f (x) = b^{x}$ (where $b > 0$ and $\neq = 1$ ), its derivative is:

$\frac{d}{d x} (b^{x}) = b^{x} \cdot ln (b)$

In other words, just rewrite the function and multiply by the natural log of the base value.

Example

1. Determine the slope of the tangent line to $f (x) = 7^{x}$ at $x = 0$ .

Solution

(spoiler)

Think derivative with the keywords “slope of the tangent line.”

$f^{'} (x) = 7^{x} \cdot ln (7)$

At $x = 0$ , the slope of the tangent line is $f^{'} (0) = 7^{0} ln (7) = ln (7)$ .

2. Where on the graph of

$f (x) = x^{2} e^{x}$

will there be horizontal tangent lines?

Solution

(spoiler)

Answer: At $(0, 0)$ and $(- 2, 4 e^{- 2})$

Using the product rule,

$f^{'} (x) = x^{2} (e^{x}) + e^{x} (2 x)$

A horizontal tangent line means the derivative $= 0$ at that point. Solve for:

$x^{2} e^{x} + 2 x e^{x} = 0$

Make sure to avoid simply moving one of the terms to the other side and dividing both sides by $x e^{x}$ , since that may eliminate a solution. Instead, always factor:

$x e^{x} (x + 2) = 0$

The graph of $y = e^{x}$ is always above the $x$ -axis and there is no solution for which $e^{x} = 0$ . So the two solutions are

$x = 0, - 2$

with corresponding $y$ -coordinates of

$f (0) = (0)^{2} \cdot e^{0} = 0$

$f (- 2) = (- 2)^{2} \cdot e^{- 2} = 4 e^{- 2} \approx 0.544$

Then the two points on $f (x)$ that have a horizontal tangent line are

$(0, 0) and (- 2, 4 e^{- 2})$

Sidenote

Power rule vs. derivatives of exponentials

Be careful not to confuse these two! Given $x$ is a variable and $b$ is a constant:

Power rule if $f (x) = x^{b}$ :
- e.g. $\frac{d}{d x} x^{3} = 3 x^{2}$
Rule for exponentials if $f (x) = b^{x}$
- e.g. $\frac{d}{d x} 3^{x} = 3^{x} ln (3)$

Logarithmic functions

For $f (x) = ln (x)$ :

$\frac{d}{d x} ln (x) = \frac{1}{x}$

Any logarithmic function $f (x) = lo g_{b} x$ (where the base $b > 0$ and $\neq = 1$ ) can be rewritten using the change-of-base rule into

$f (x) = \frac{ln ( x )}{ln ( b )}$

$= \frac{1}{ln ( b )} \cdot ln (x)$

$\frac{1}{ln ( b )}$ is pulled to. the front because it’s a constant. Then the derivative is

$\frac{d}{d x} lo g_{b} x = \frac{1}{x \cdot ln ( b )}$

Notice that $lo g_{e} (x) = ln (x)$ and the derivative is $\frac{1}{x \cdot ln ( e )} = \frac{1}{x}$ .

Sidenote

On the derivative's domain

The derivative is only valid in the domain of the original function. For all logarithmic functions, the argument must be $> 0$ .

Examples

2. Let $f (x) = lo g_{1/2} (x)$ . Determine for which values of $x$ the slope of the tangent line to $f (x)$ is positive and the values for which it is negative.

(spoiler)

Answer: Negative slope for all $x$ -values in $(0, \infty)$ .

$f^{'} (x) = \frac{1}{x \cdot ln ( 1/2 )}$

$= \frac{1}{ln ( 1/2 )} \cdot \frac{1}{x}$

$\approx - 1.44 \cdot \frac{1}{x}$

Graphing $lo g_{1/2} (x)$ confirms the function is always decreasing as we move from left to right.

3. Find the derivative of

$f (x) = lo g_{x} (4)$

Solution

(spoiler)

Using the change of base formula,

$f (x) = \frac{ln ( 4 )}{ln ( x )}$

Differentiating with the quotient rule,

$f^{'} (x) = \frac{ln ( x ) ( 0 ) - ln ( 4 ) \cdot \frac{1}{x}}{( ln ( x ) ) ^{2}}$

$= - \frac{ln ( 4 )}{x ( ln ( x ) ) ^{2}}$

Here’s a way to confirm that this is the derivative in Desmos:

1. Type

$\frac{d}{d x} (the original function)$

2. Type in the derivative function we found. Use as many parentheses as possible - it’s better to err on the side of too many than not enough.

3. Check that the graphs match.

Trigonometric functions

Let’s start with the basic trig functions: $sin (x)$ and $cos (x)$ .

$\frac{d}{d x} sin (x) = cos (x)$

$\frac{d}{d x} cos (x) = - sin (x)$

From there, the derivative of $tan (x)$ can be derived using the quotient rule (or product rule, if so inclined) by rewriting $tan (x)$ as $\frac{sin ( x )}{cos ( x )}$ .

1. Find the derivative of $f (x) = tan (x)$ by first rewriting it as a quotient.

Solution

(spoiler)

Answer: $sec^{2} (x)$

Using the quotient rule on $f (x) = \frac{sin ( x )}{cos ( x )}$ ,

$f^{'} (x) = \frac{cos ( x ) \cdot cos ( x ) - sin ( x ) \cdot ( - sin ( x ))}{cos ^{2} ( x )}$

$= \frac{cos ^{2} ( x ) + sin ^{2} ( x )}{cos ^{2} ( x )}$

Using the Pythagorean identity
$sin^{2} (x) + cos^{2} (x) = 1$ ,

$f^{'} (x) = \frac{1}{cos ^{2} ( x )} = sec^{2} (x)$

Let’s do the same for the derivative of $cot (x)$ .

2. Find the derivative of $f (x) = cot (x)$ by rewriting it as a quotient.

Solution

(spoiler)

Answer: $- csc^{2} (x)$

Using the quotient rule on $f (x) = \frac{cos ( x )}{sin ( x )}$ ,

$f^{'} (x) = \frac{sin ( x ) \cdot ( - sin ( x )) - cos ( x ) \cdot cos ( x )}{sin ^{2} ( x )}$

$= \frac{- sin ^{2} ( x ) - cos ^{2} ( x )}{sin ^{2} ( x )}$

$= \frac{- ( sin ^{2} ( x ) + cos ^{2} ( x ))}{sin ^{2} ( x )}$

$= \frac{- 1}{sin ^{2} ( x )} = - csc^{2} (x)$

Let’s do the same for the remaining two reciprocal functions:

3. Find the derivatives of

a. $a (x) = sec (x)$
b. $b (x) = csc (x)$

Solutions

(spoiler)

a. Since $sec (x) = \frac{1}{cos ( x )},$

$a^{'} (x) = \frac{cos ( x ) ( 0 ) - 1 ( - sin ( x ))}{cos ^{2} ( x )}$

$= \frac{sin ( x )}{cos ^{2} ( x )}$

$= \frac{1}{cos ( x )} \cdot \frac{sin ( x )}{cos ( x )}$

$= sec (x) tan (x)$

b. Since $csc (x) = \frac{1}{sin ( x )},$

$a^{'} (x) = \frac{sin ( x ) ( 0 ) - 1 ( cos ( x ))}{sin ^{2} ( x )}$

$= \frac{- cos ( x )}{sin ^{2} ( x )}$

$= \frac{- 1}{sin ( x )} \cdot \frac{cos ( x )}{sin ( x )}$

$= - csc (x) cot (x)$

Memorizing the trig derivatives will save time but it’s also handy to know how to derive them from the derivatives of $sin (x)$ and $cos (x)$ .

Key points

These tables summarize the special derivatives:

For exponential and logarithmic functions:

$f (x)$	$f^{'} (x)$
$e^{x}$	$e^{x}$
$a^{x}$	$a^{x} ln (a)$
$ln (x)$	$\frac{1}{x}$
$lo g_{b} (x)$	$\frac{1}{x ln ( b )}$

For trig functions:

$f (x)$	$f^{'} (x)$
$sin (x)$	$cos (x)$
$tan (x)$	$sec^{2} (x)$
$cot (x)$	$- csc^{2} (x)$
$sec (x)$	$sec (x) tan (x)$
$csc (x)$	$- csc (x) cot (x)$