2.5 Special derivatives

Achievable AP Calculus AB

2. Derivative basics

Our AP Calculus AB course is currently in development and is a work-in-progress.

Special derivatives

6 min read

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What you’ll learn:

Derivatives of exponential, logarithmic, and trigonometry functions

So far, we’ve worked with derivatives of polynomials and rational functions. Many important functions don’t fit those categories, though. Here, you’ll learn the special derivative rules for exponential, logarithmic, and trigonometric functions.

Exponential functions

Exponential functions model rapid growth and decay. They show up in population models, radioactive decay, and compound interest.

The most important exponential function is the natural exponential function

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

It has a key property: its rate of change is equal to itself.

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

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

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

In other words, you rewrite the original function and multiply by $ln (b)$ .

Example

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

Solution

(spoiler)

“Slope of the tangent line” means “evaluate the derivative at that $x$ -value.”

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

At $x = 0$ , the slope is

$f^{'} (0) = 7^{0} ln (7) = ln (7) .$

Where on the graph of

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

will there be horizontal tangent lines?

Solution

(spoiler)

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

Use the product rule:

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

Horizontal tangent lines occur where $f^{'} (x) = 0$ , so solve

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

Avoid dividing both sides by $x e^{x}$ right away, since that can remove solutions. Instead, factor:

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

Because $e^{x} \neq = 0$ for any real $x$ , the solutions come from

$x = 0 or x + 2 = 0 \Rightarrow x = - 2$

Now find the corresponding $y$ -values:

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

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

So the points are

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

Sidenote

Power rule vs. derivatives of exponentials

Be careful not to mix these up. Here, $x$ 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 the natural logarithm $f (x) = ln (x)$ ,

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

Any logarithmic function

$f (x) = lo g_{b} x$ (where the base $b > 0$ and $b \neq = 1$ )

can be rewritten using the change-of-base rule:

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

Since $\frac{1}{ln ( b )}$ is a constant, it stays out front when you differentiate. That gives

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

Notice that $lo g_{e} (x) = ln (x)$ , and this formula becomes

$\frac{1}{x ln ( e )} = \frac{1}{x} .$

Sidenote

On the derivative's domain

A derivative formula is only valid where the original function is defined. For logarithms, the argument must be $> 0$ .

Examples

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 $x > 0$ , the domain of $f^{'} (x)$ is also $x > 0$ . For every positive input $x$ , the factor $\frac{1}{x}$ is positive, while $\frac{1}{l n ( 1/2 )}$ is negative. So $f^{'} (x)$ is always negative.

Graphing $lo g_{1/2} (x)$ confirms the function decreases as you move from left to right.

Find the derivative of

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

Solution

(spoiler)

Start by rewriting with change of base:

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

Now differentiate using the quotient rule. Here, $ln (4)$ is a constant, so its derivative is $0$ .

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

Here’s one way to confirm this derivative in Desmos:

Type

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

Type the derivative function you found. Use plenty of parentheses.
Check that the graphs match.

Trigonometric functions

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, you can derive other trig derivatives by rewriting them in terms of $sin (x)$ and $cos (x)$ .

For example, you can find the derivative of $tan (x)$ by rewriting it as a quotient:

$tan (x) = \frac{sin ( x )}{cos ( x )} .$

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 $sin^{2} (x) + cos^{2} (x) = 1$ ,

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

You can do the same for $cot (x)$ by rewriting it as

$cot (x) = \frac{cos ( x )}{sin ( x )} .$

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 $f (x) = \frac{cos ( x )}{sin ( x )}$ ,

$f^{'} (x) = \frac{sin ( x ) \cdot ( - sin ( x )) - cos ( x ) \cdot cos ( x )}{sin ^{2} ( x )} = \frac{- sin ^{2} ( x ) - cos ^{2} ( x )}{sin ^{2} ( x )} = \frac{- ( sin ^{2} ( x ) + cos ^{2} ( x ))}{sin ^{2} ( x )} = \frac{- 1}{sin ^{2} ( x )} = - csc^{2} (x)$

Now do the same for the remaining two reciprocal functions:

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 saves time, but it’s also useful to know how to derive them from the derivatives of $sin (x)$ and $cos (x)$ .

Exponential functions

Derivative of $e^{x}$ is $e^{x}$
Derivative of $b^{x}$ is $b^{x} ln (b)$
Power rule vs. exponential rule:
- Power rule: $x^{b} \to b x^{b - 1}$
- Exponential: $b^{x} \to b^{x} ln (b)$

Logarithmic functions

Derivative of $ln (x)$ is $\frac{1}{x}$
Derivative of $lo g_{b} (x)$ is $\frac{1}{x l n ( b )}$
Derivative only defined for $x > 0$

Trigonometric functions

$\frac{d}{d x} sin (x) = cos (x)$
$\frac{d}{d x} cos (x) = - sin (x)$
$\frac{d}{d x} tan (x) = sec^{2} (x)$
$\frac{d}{d x} cot (x) = - csc^{2} (x)$
$\frac{d}{d x} sec (x) = sec (x) tan (x)$
$\frac{d}{d x} csc (x) = - csc (x) cot (x)$

Special derivatives

What you’ll learn:

Derivatives of exponential, logarithmic, and trigonometry functions

Exponential functions

Exponential functions model rapid growth and decay. They show up in population models, radioactive decay, and compound interest.

The most important exponential function is the natural exponential function

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

It has a key property: its rate of change is equal to itself.

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

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

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

In other words, you rewrite the original function and multiply by $ln (b)$ .

Example

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

Solution

(spoiler)

“Slope of the tangent line” means “evaluate the derivative at that $x$ -value.”

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

At $x = 0$ , the slope is

$f^{'} (0) = 7^{0} ln (7) = ln (7) .$

Where on the graph of

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

will there be horizontal tangent lines?

Solution

(spoiler)

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

Use the product rule:

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

Horizontal tangent lines occur where $f^{'} (x) = 0$ , so solve

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

Avoid dividing both sides by $x e^{x}$ right away, since that can remove solutions. Instead, factor:

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

Because $e^{x} \neq = 0$ for any real $x$ , the solutions come from

$x = 0 or x + 2 = 0 \Rightarrow x = - 2$

Now find the corresponding $y$ -values:

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

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

So the points are

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

Sidenote

Power rule vs. derivatives of exponentials

Be careful not to mix these up. Here, $x$ 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 the natural logarithm $f (x) = ln (x)$ ,

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

Any logarithmic function

$f (x) = lo g_{b} x$ (where the base $b > 0$ and $b \neq = 1$ )

can be rewritten using the change-of-base rule:

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

Since $\frac{1}{ln ( b )}$ is a constant, it stays out front when you differentiate. That gives

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

Notice that $lo g_{e} (x) = ln (x)$ , and this formula becomes

$\frac{1}{x ln ( e )} = \frac{1}{x} .$

Sidenote

On the derivative's domain

A derivative formula is only valid where the original function is defined. For logarithms, the argument must be $> 0$ .

Examples

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 decreases as you move from left to right.

Find the derivative of

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

Solution

(spoiler)

Start by rewriting with change of base:

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

Now differentiate using the quotient rule. Here, $ln (4)$ is a constant, so its derivative is $0$ .

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

Here’s one way to confirm this derivative in Desmos:

Type

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

Type the derivative function you found. Use plenty of parentheses.
Check that the graphs match.

Trigonometric functions

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, you can derive other trig derivatives by rewriting them in terms of $sin (x)$ and $cos (x)$ .

For example, you can find the derivative of $tan (x)$ by rewriting it as a quotient:

$tan (x) = \frac{sin ( x )}{cos ( x )} .$

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 $sin^{2} (x) + cos^{2} (x) = 1$ ,

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

You can do the same for $cot (x)$ by rewriting it as

$cot (x) = \frac{cos ( x )}{sin ( x )} .$

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 $f (x) = \frac{cos ( x )}{sin ( x )}$ ,

Now do the same for the remaining two reciprocal functions:

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 saves time, but it’s also useful to know how to derive them from the derivatives of $sin (x)$ and $cos (x)$ .

Achievable AP Calculus AB

2. Derivative basics

Our AP Calculus AB course is currently in development and is a work-in-progress.

Special derivatives

6 min read

Font

Discuss

Feedback

What you’ll learn:

Derivatives of exponential, logarithmic, and trigonometry functions

Exponential functions

Exponential functions model rapid growth and decay. They show up in population models, radioactive decay, and compound interest.

The most important exponential function is the natural exponential function

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

It has a key property: its rate of change is equal to itself.

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

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

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

In other words, you rewrite the original function and multiply by $ln (b)$ .

Example

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

Solution

(spoiler)

“Slope of the tangent line” means “evaluate the derivative at that $x$ -value.”

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

At $x = 0$ , the slope is

$f^{'} (0) = 7^{0} ln (7) = ln (7) .$

Where on the graph of

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

will there be horizontal tangent lines?

Solution

(spoiler)

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

Use the product rule:

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

Horizontal tangent lines occur where $f^{'} (x) = 0$ , so solve

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

Avoid dividing both sides by $x e^{x}$ right away, since that can remove solutions. Instead, factor:

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

Because $e^{x} \neq = 0$ for any real $x$ , the solutions come from

$x = 0 or x + 2 = 0 \Rightarrow x = - 2$

Now find the corresponding $y$ -values:

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

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

So the points are

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

Sidenote

Power rule vs. derivatives of exponentials

Be careful not to mix these up. Here, $x$ 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 the natural logarithm $f (x) = ln (x)$ ,

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

Any logarithmic function

$f (x) = lo g_{b} x$ (where the base $b > 0$ and $b \neq = 1$ )

can be rewritten using the change-of-base rule:

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

Since $\frac{1}{ln ( b )}$ is a constant, it stays out front when you differentiate. That gives

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

Notice that $lo g_{e} (x) = ln (x)$ , and this formula becomes

$\frac{1}{x ln ( e )} = \frac{1}{x} .$

Sidenote

On the derivative's domain

A derivative formula is only valid where the original function is defined. For logarithms, the argument must be $> 0$ .

Examples

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 decreases as you move from left to right.

Find the derivative of

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

Solution

(spoiler)

Start by rewriting with change of base:

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

Now differentiate using the quotient rule. Here, $ln (4)$ is a constant, so its derivative is $0$ .

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

Here’s one way to confirm this derivative in Desmos:

Type

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

Type the derivative function you found. Use plenty of parentheses.
Check that the graphs match.

Trigonometric functions

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, you can derive other trig derivatives by rewriting them in terms of $sin (x)$ and $cos (x)$ .

For example, you can find the derivative of $tan (x)$ by rewriting it as a quotient:

$tan (x) = \frac{sin ( x )}{cos ( x )} .$

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 $sin^{2} (x) + cos^{2} (x) = 1$ ,

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

You can do the same for $cot (x)$ by rewriting it as

$cot (x) = \frac{cos ( x )}{sin ( x )} .$

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 $f (x) = \frac{cos ( x )}{sin ( x )}$ ,

Now do the same for the remaining two reciprocal functions:

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 saves time, but it’s also useful to know how to derive them from the derivatives of $sin (x)$ and $cos (x)$ .

Exponential functions

Derivative of $e^{x}$ is $e^{x}$
Derivative of $b^{x}$ is $b^{x} ln (b)$
Power rule vs. exponential rule:
- Power rule: $x^{b} \to b x^{b - 1}$
- Exponential: $b^{x} \to b^{x} ln (b)$

Logarithmic functions

Derivative of $ln (x)$ is $\frac{1}{x}$
Derivative of $lo g_{b} (x)$ is $\frac{1}{x l n ( b )}$
Derivative only defined for $x > 0$

Trigonometric functions

$\frac{d}{d x} sin (x) = cos (x)$
$\frac{d}{d x} cos (x) = - sin (x)$
$\frac{d}{d x} tan (x) = sec^{2} (x)$
$\frac{d}{d x} cot (x) = - csc^{2} (x)$
$\frac{d}{d x} sec (x) = sec (x) tan (x)$
$\frac{d}{d x} csc (x) = - csc (x) cot (x)$

Special derivatives

What you’ll learn:

Derivatives of exponential, logarithmic, and trigonometry functions

Exponential functions

Exponential functions model rapid growth and decay. They show up in population models, radioactive decay, and compound interest.

The most important exponential function is the natural exponential function

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

It has a key property: its rate of change is equal to itself.

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

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

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

In other words, you rewrite the original function and multiply by $ln (b)$ .

Example

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

Solution

(spoiler)

“Slope of the tangent line” means “evaluate the derivative at that $x$ -value.”

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

At $x = 0$ , the slope is

$f^{'} (0) = 7^{0} ln (7) = ln (7) .$

Where on the graph of

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

will there be horizontal tangent lines?

Solution

(spoiler)

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

Use the product rule:

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

Horizontal tangent lines occur where $f^{'} (x) = 0$ , so solve

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

Avoid dividing both sides by $x e^{x}$ right away, since that can remove solutions. Instead, factor:

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

Because $e^{x} \neq = 0$ for any real $x$ , the solutions come from

$x = 0 or x + 2 = 0 \Rightarrow x = - 2$

Now find the corresponding $y$ -values:

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

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

So the points are

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

Sidenote

Power rule vs. derivatives of exponentials

Be careful not to mix these up. Here, $x$ 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 the natural logarithm $f (x) = ln (x)$ ,

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

Any logarithmic function

$f (x) = lo g_{b} x$ (where the base $b > 0$ and $b \neq = 1$ )

can be rewritten using the change-of-base rule:

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

Since $\frac{1}{ln ( b )}$ is a constant, it stays out front when you differentiate. That gives

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

Notice that $lo g_{e} (x) = ln (x)$ , and this formula becomes

$\frac{1}{x ln ( e )} = \frac{1}{x} .$

Sidenote

On the derivative's domain

A derivative formula is only valid where the original function is defined. For logarithms, the argument must be $> 0$ .

Examples

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 decreases as you move from left to right.

Find the derivative of

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

Solution

(spoiler)

Start by rewriting with change of base:

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

Now differentiate using the quotient rule. Here, $ln (4)$ is a constant, so its derivative is $0$ .

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

Here’s one way to confirm this derivative in Desmos:

Type

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

Type the derivative function you found. Use plenty of parentheses.
Check that the graphs match.

Trigonometric functions

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, you can derive other trig derivatives by rewriting them in terms of $sin (x)$ and $cos (x)$ .

For example, you can find the derivative of $tan (x)$ by rewriting it as a quotient:

$tan (x) = \frac{sin ( x )}{cos ( x )} .$

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 $sin^{2} (x) + cos^{2} (x) = 1$ ,

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

You can do the same for $cot (x)$ by rewriting it as

$cot (x) = \frac{cos ( x )}{sin ( x )} .$

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 $f (x) = \frac{cos ( x )}{sin ( x )}$ ,

Now do the same for the remaining two reciprocal functions:

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 saves time, but it’s also useful to know how to derive them from the derivatives of $sin (x)$ and $cos (x)$ .