What you’ll learn:

How to differentiate composite functions with the chain rule
Applying the chain rule with logarithmic, exponential, trig, and power functions
Combining the chain rule with product and quotient rules

If you want the derivative of a function like $h (x) = (2 x + 1)^{3}$ , you could expand it and use the power rule term-by-term, or rewrite it as a product of three identical factors and use the product rule. Both approaches work, but they get tedious quickly (especially with larger exponents, more complicated polynomials, or fractional powers).

A faster method is designed specifically for composite functions - functions built by plugging one function into another. That method is the chain rule.

Chain rule:

$\frac{d}{d x} [f (g (x))] = f^{'} (g (x)) \cdot g^{'} (x)$

Here’s the idea in words:

Differentiate the outer function, keeping the inner function intact.
Then multiply by the derivative of the inner function.

Another common way to write the chain rule uses Leibniz notation:

Let $y = f (g (x))$ and $u = g (x)$ .

Then $y = f (u)$ and

$\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}$

One way to remember this form is to notice that the $d u$ factors “cancel” visually, leaving $\frac{d y}{d x}$ . The idea is the same: differentiate the outer function and then multiply by the derivative of the inner function.

Examples

1. Differentiate

$y = (2 x + 1)^{3}$

Solution

$y$ is a composite function, so we identify the inner and outer parts. Below, the inner function is boxed:

$outer (2 x + 1 inner)^{3}$

For clarity:

Outer function: $f (u) = u^{3}$
Inner function: $g (x) = 2 x + 1$

Differentiate the outer function (power rule), keeping the inner function intact:

$f^{'} (u) = 3 u^{2}$

$f^{'} (g (x)) = 3 (2 x + 1)^{2}$

Differentiate the inner function:

$g^{'} (x) = 2$

Multiply the results (chain rule):

$y^{'} = f^{'} (g (x)) \cdot g^{'} (x) = 3 (2 x + 1)^{2} \cdot 2 = 6 (2 x + 1)^{2}$

Let’s try one with the natural logarithm. Recall from section 2.5 that $\frac{d}{d x} [ln (x)] = \frac{1}{x}$ .

2. Differentiate

$h (x) = ln (3 x^{2} + x)$

Solution

(spoiler)

$h (x)$ is a composite function $f (g (x))$ with:

$outer ln (3 x^{2} + x inner)$

For clarity:

Outer function: $f (u) = ln (u)$
Inner function: $g (x) = 3 x^{2} + x$

Differentiate the outer function, keeping the inner function intact:

$f^{'} (u) = \frac{1}{u}$

$f^{'} (g (x)) = \frac{1}{3 x ^{2} + x}$

Differentiate the inner function:

$g^{'} (x) = 6 x + 1$

Multiply the results (chain rule):

$h^{'} (x) = f^{'} (g (x)) \cdot g^{'} (x) = \frac{6 x + 1}{3 x ^{2} + x}$

Similarly, recall that for exponential functions,

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

and

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

Let’s do 2 examples that use the chain rule.

3. Differentiate:

a) $a (x) = e^{- 2 x}$

b) $b (x) = 2^{3 x + 1}$

Solutions

a) $a (x) = e^{- 2 x}$

(spoiler)

Outer function: $f (u) = e^{u}$
Inner function: $g (x) = - 2 x$

Differentiate the outer function:

$f^{'} (u) = e^{u}$

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

Differentiate the inner function:

$g^{'} (x) = - 2$

Multiply:

$a^{'} (x) = f^{'} (g (x)) \cdot g^{'} (x) = - 2 e^{- 2 x}$

b) $b (x) = 2^{3 x + 1}$

(spoiler)

Outer function: $f (u) = 2^{u}$
Inner function: $g (x) = 3 x + 1$

Differentiate the outer function:

$f^{'} (u) = 2^{u} \cdot l n (2)$

$f^{'} (g (x)) = 2^{3 x + 1} \cdot ln (2)$

Differentiate the inner function:

$g^{'} (x) = 3$

Multiply:

$b^{'} (x) = f^{'} (g (x)) \cdot g^{'} (x) = (2^{3 x + 1} \cdot ln (2)) \cdot (3) = (3 ln (2)) \cdot 2^{3 x + 1}$

Combining rules

The chain rule often appears together with other derivative rules. The main thing to watch is order: decide which overall structure you have first (product, quotient, or composition), then apply the needed rules inside that structure.

For example, the next problem uses the product rule first, and then uses the chain rule while differentiating one of the factors.

3. Differentiate

$h (x) = 3 x (x^{2} - 1)^{3}$

Solution

(spoiler)

$h (x)$ is a product of two functions:

$f (x) = 3 x$
$g (x) = (x^{2} - 1)^{3}$

Notice that $g (x)$ is composite, so the chain rule is needed to find $g^{'} (x)$ .

Function	Expression
$f (x)$	$3 x$
$f^{'} (x)$	$3$
$g (x)$	$(x^{2} - 1)^{3}$
$g^{'} (x)$	$3 (x^{2} - 1)^{2} \cdot 2 x$

Putting these into the product rule:

$h^{'} (x) = 3 x \cdot [3 (x^{2} - 1)^{2} \cdot 2 x] + 3 \cdot (x^{2} - 1)^{3}$

$= 18 x^{2} (x^{2} - 1)^{2} + 3 (x^{2} - 1)^{3}$

To simplify, factor out the common factors $3$ and $(x^{2} - 1)^{2}$ :

$= 3 (x^{2} - 1)^{2} [6 x^{2} + (x^{2} - 1)]$

$= 3 (x^{2} - 1)^{2} (7 x^{2} - 1)$

Contrast that with the next problem. Here, it’s most natural to use the power rule (with the chain rule) first, and then use the product rule when differentiating the inner function.

4. Differentiate

$h (x) = (e^{x} sin (x))^{2}$

Solution

(spoiler)

$h (x)$ is a composite function with

$outer (e^{x} sin (x) inner)^{2}$

Differentiate the outer function:

$f^{'} (g (x)) = 2 (g (x) e^{x} sin (x))^{1}$

Differentiate the inner function:

The inner function $g (x) = e^{x} sin (x)$ is a product, so apply the product rule:

$g^{'} (x) = e^{x} cos (x) + e^{x} sin (x) = e^{x} [cos (x) + sin (x)]$

Multiply:

$h^{'} (x) = 2 e^{x} sin (x) \cdot e^{x} [cos (x) + sin (x)] = 2 e^{2 x} sin (x) [cos (x) + sin (x)]$

5. Differentiate

$h (x) = \frac{x}{2 ( x ^{2} + 1 ) ^{2}}$

Solution

(spoiler)

For easier algebra, pull out the constant factor $\frac{1}{2}$ :

$h (x) = \frac{1}{2} \cdot \frac{x}{( x ^{2} + 1 ) ^{2}}$

Now $h (x)$ is a quotient of two functions:

$f (x) = x$
$g (x) = (x^{2} + 1)^{2}$

Use the quotient rule. Note that $g (x)$ is composite, so the chain rule is needed for $g^{'} (x)$ .

Function	Expression
$f (x)$	$x$
$f^{'} (x)$	$1$
$g (x)$	$(x^{2} + 1)^{2}$
$g^{'} (x)$	$2 (x^{2} + 1) \cdot 2 x$

Putting the pieces into the quotient rule:

$h^{'} (x) = \frac{1}{2} \cdot \frac{( x ^{2} + 1 ) ^{2} ( 1 ) - x \cdot 2 ( x ^{2} + 1 ) \cdot 2 x}{( x ^{2} + 1 ) ^{4}}$

$= \frac{( x ^{2} + 1 ) ^{2} - 4 x ^{2} ( x ^{2} + 1 )}{2 ( x ^{2} + 1 ) ^{4}}$

$= \frac{( x ^{2} + 1 ) ( x ^{2} + 1 - 4 x ^{2} )}{2 ( x ^{2} + 1 ) ^{4}}$

$= \frac{1 - 3 x ^{2}}{2 ( x ^{2} + 1 ) ^{3}}$

You can also rewrite a quotient as a product and then use the product rule, if that feels faster. Using the same function as above,

5b. First turn

$h (x) = \frac{x}{2 ( x ^{2} + 1 ) ^{2}}$

into a product of two functions:

$h (x) = \frac{1}{2} x (x^{2} + 1)^{- 2}$

and then find the derivative using the product rule.

Solution

(spoiler)

As before, keep the constant multiple $\frac{1}{2}$ out in front:

$h (x) = \frac{1}{2} x (x^{2} + 1)^{- 2}$

The remaining part is a product of:

$f (x) = x$
$g (x) = (x^{2} + 1)^{- 2}$

Use the product rule. Since $g (x)$ is composite, use the chain rule to find $g^{'} (x)$ .

Function	Expression
$f (x)$	$x$
$f^{'} (x)$	$1$
$g (x)$	$(x^{2} + 1)^{- 2}$
$g^{'} (x)$	$- 2 (x^{2} + 1)^{- 3} \cdot 2 x$

Now apply the product rule:

$h^{'} (x) = \frac{1}{2} \cdot [x (- 4 x (x^{2} + 1)^{- 3}) + 1 (x^{2} + 1)^{- 2}]$

$= \frac{1}{2} \cdot [\frac{- 4 x ^{2}}{( x ^{2} + 1 ) ^{3}} + \frac{1}{( x ^{2} + 1 ) ^{2}}]$

$= \frac{1}{2} \cdot [\frac{- 4 x ^{2}}{( x ^{2} + 1 ) ^{3}} + \frac{( x ^{2} + 1 )}{( x ^{2} + 1 ) ^{3}}]$

$= \frac{1 - 3 x ^{2}}{2 ( x ^{2} + 1 ) ^{3}}$

Some questions ask you to apply derivative rules using only a table of values:

6. $f (x), g (x),$ and their derivatives have the following values shown in the table:

$x$ $f (x)$ $g (x)$ $f^{'} (x)$ $g^{'} (x)$

$1$ $3$ $2$ $- 1$ $5$

$2$ $0.2$ $0.5$ $5$ $3$

$3$ $- 1$ $1$ $4$ $- 3$

Evaluate the first derivative of the function

$h (x) = f (g (x) - x^{2})$

at $x = 1$ .

$x$	$f (x)$	$g (x)$	$f^{'} (x)$	$g^{'} (x)$
$1$	$3$	$2$	$- 1$	$5$
$2$	$0.2$	$0.5$	$5$	$3$
$3$	$- 1$	$1$	$4$	$- 3$

Solution

(spoiler)

Because $h (x) = f (g (x) - x^{2})$ is a composite function, apply the chain rule:

$\frac{d}{d x} [f (g (x) - x^{2})]$

$= f^{'} (g (x) - x^{2}) \cdot (g^{'} (x) - 2 x)$

Now evaluate at $x = 1$ :

$h^{'} (1) = f^{'} (g (1) - 1^{2}) \cdot (g^{'} (1) - 2 (1))$

$= f^{'} (2 - 1) \cdot (5 - 2)$

$= f^{'} (1) \cdot 3$

$= - 1 \cdot 3$

$= - 3$

Challenge problems

1. Differentiate $h (t) = 3 2 π - π t$ .

Solution

(spoiler)

Rewrite the square root as a power:

$h (t) = 3 (2 π - π t)^{1/2}$

The composite function is:

$3 (inner 2 π - π t)^{1/2} outer$

Outer: $f (u) = 3 u^{1/2}$
Inner: $g (t) = 2 π - π t$

Differentiate the outer function:

$f^{'} (u) = \frac{3}{2} u^{- 1/2}$

$f^{'} (g (t)) = \frac{3}{2} (2 π - π t)^{- 1/2}$

Differentiate the inner function:

( $2 π$ is a constant, so its derivative is $0$ . The derivative of $- π t$ is $- π$ )

$g^{'} (t) = - π$

Multiply:

$h^{'} (t) = \frac{3}{2} (2 π - π t)^{- 1/2} (- π)$

$= - \frac{3 π}{2 2 π - π t}$

2. Differentiate $h (x) = \frac{e ^{x^{2}}}{e ^{c o s (x)}}$

Solution

(spoiler)

Instead of using the quotient rule, rewrite the expression using $\frac{a ^{m}}{a ^{n}} = a^{m - n}$ :

$h (x) = e^{(x^{2} - c o s (x))}$

Outer: $f (u) = e^{u}$
Inner: $g (x) = x^{2} - cos (x)$

Differentiate the outer:

$f^{'} (u) = e^{u}$

$f^{'} (g (x)) = e^{x^{2} - c o s (x)}$

Differentiate the inner:

$g^{'} (x) = 2 x + sin (x)$

Multiply:

$h^{'} (x) = e^{(x^{2} - c o s (x))} \cdot (2 x + sin (x))$

This can also be written as

$h^{'} (x) = \frac{e ^{x^{2}} ( 2 x + sin ( x ))}{e ^{c o s (x)}}$

3. Differentiate $A (x) = sin^{3} (ln x)$ .

Solution

(spoiler)

First, rewrite $sin^{3} (ln x)$ as $(sin (ln x))^{3}$ :

$A (x) = (sin (ln x))^{3}$

This function has multiple layers of composition and requires two applications of the chain rule:

Outer: $f (u) = u^{3}$
Middle: $u = sin (v)$
Inner: $v = ln (x)$

Differentiate the outer:

$f^{'} (u) = 3 u^{2}$

Differentiate the middle:

$u^{'} = cos (v)$

Differentiate the inner:

$v^{'} = \frac{1}{x}$

Apply the chain rule layer by layer:

$A^{'} (x) = der. of outer 3 (sin (ln x))^{2} \times cos (ln x) der. of middle \times der. of inner \frac{1}{x}$

$= \frac{3 sin ^{2} ( ln x ) \cdot cos ( ln x )}{x}$

What you’ll learn:

How to differentiate composite functions with the chain rule
Applying the chain rule with logarithmic, exponential, trig, and power functions
Combining the chain rule with product and quotient rules

A faster method is designed specifically for composite functions - functions built by plugging one function into another. That method is the chain rule.

Chain rule:

$\frac{d}{d x} [f (g (x))] = f^{'} (g (x)) \cdot g^{'} (x)$

Here’s the idea in words:

Differentiate the outer function, keeping the inner function intact.
Then multiply by the derivative of the inner function.

Another common way to write the chain rule uses Leibniz notation:

Let $y = f (g (x))$ and $u = g (x)$ .

Then $y = f (u)$ and

$\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}$

Examples

1. Differentiate

$y = (2 x + 1)^{3}$

Solution

$y$ is a composite function, so we identify the inner and outer parts. Below, the inner function is boxed:

$outer (2 x + 1 inner)^{3}$

For clarity:

Outer function: $f (u) = u^{3}$
Inner function: $g (x) = 2 x + 1$

Differentiate the outer function (power rule), keeping the inner function intact:

$f^{'} (u) = 3 u^{2}$

$f^{'} (g (x)) = 3 (2 x + 1)^{2}$

Differentiate the inner function:

$g^{'} (x) = 2$

Multiply the results (chain rule):

$y^{'} = f^{'} (g (x)) \cdot g^{'} (x) = 3 (2 x + 1)^{2} \cdot 2 = 6 (2 x + 1)^{2}$

Let’s try one with the natural logarithm. Recall from section 2.5 that $\frac{d}{d x} [ln (x)] = \frac{1}{x}$ .

2. Differentiate

$h (x) = ln (3 x^{2} + x)$

Solution

(spoiler)

$h (x)$ is a composite function $f (g (x))$ with:

$outer ln (3 x^{2} + x inner)$

For clarity:

Outer function: $f (u) = ln (u)$
Inner function: $g (x) = 3 x^{2} + x$

Differentiate the outer function, keeping the inner function intact:

$f^{'} (u) = \frac{1}{u}$

$f^{'} (g (x)) = \frac{1}{3 x ^{2} + x}$

Differentiate the inner function:

$g^{'} (x) = 6 x + 1$

Multiply the results (chain rule):

$h^{'} (x) = f^{'} (g (x)) \cdot g^{'} (x) = \frac{6 x + 1}{3 x ^{2} + x}$

Similarly, recall that for exponential functions,

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

and

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

Let’s do 2 examples that use the chain rule.

3. Differentiate:

a) $a (x) = e^{- 2 x}$

b) $b (x) = 2^{3 x + 1}$

Solutions

a) $a (x) = e^{- 2 x}$

(spoiler)

Outer function: $f (u) = e^{u}$
Inner function: $g (x) = - 2 x$

Differentiate the outer function:

$f^{'} (u) = e^{u}$

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

Differentiate the inner function:

$g^{'} (x) = - 2$

Multiply:

$a^{'} (x) = f^{'} (g (x)) \cdot g^{'} (x) = - 2 e^{- 2 x}$

b) $b (x) = 2^{3 x + 1}$

(spoiler)

Outer function: $f (u) = 2^{u}$
Inner function: $g (x) = 3 x + 1$

Differentiate the outer function:

$f^{'} (u) = 2^{u} \cdot l n (2)$

$f^{'} (g (x)) = 2^{3 x + 1} \cdot ln (2)$

Differentiate the inner function:

$g^{'} (x) = 3$

Multiply:

$b^{'} (x) = f^{'} (g (x)) \cdot g^{'} (x) = (2^{3 x + 1} \cdot ln (2)) \cdot (3) = (3 ln (2)) \cdot 2^{3 x + 1}$

Combining rules

For example, the next problem uses the product rule first, and then uses the chain rule while differentiating one of the factors.

3. Differentiate

$h (x) = 3 x (x^{2} - 1)^{3}$

Solution

(spoiler)

$h (x)$ is a product of two functions:

$f (x) = 3 x$
$g (x) = (x^{2} - 1)^{3}$

Notice that $g (x)$ is composite, so the chain rule is needed to find $g^{'} (x)$ .

Function	Expression
$f (x)$	$3 x$
$f^{'} (x)$	$3$
$g (x)$	$(x^{2} - 1)^{3}$
$g^{'} (x)$	$3 (x^{2} - 1)^{2} \cdot 2 x$

Putting these into the product rule:

$h^{'} (x) = 3 x \cdot [3 (x^{2} - 1)^{2} \cdot 2 x] + 3 \cdot (x^{2} - 1)^{3}$

$= 18 x^{2} (x^{2} - 1)^{2} + 3 (x^{2} - 1)^{3}$

To simplify, factor out the common factors $3$ and $(x^{2} - 1)^{2}$ :

$= 3 (x^{2} - 1)^{2} [6 x^{2} + (x^{2} - 1)]$

$= 3 (x^{2} - 1)^{2} (7 x^{2} - 1)$

Contrast that with the next problem. Here, it’s most natural to use the power rule (with the chain rule) first, and then use the product rule when differentiating the inner function.

4. Differentiate

$h (x) = (e^{x} sin (x))^{2}$

Solution

(spoiler)

$h (x)$ is a composite function with

$outer (e^{x} sin (x) inner)^{2}$

Differentiate the outer function:

$f^{'} (g (x)) = 2 (g (x) e^{x} sin (x))^{1}$

Differentiate the inner function:

The inner function $g (x) = e^{x} sin (x)$ is a product, so apply the product rule:

$g^{'} (x) = e^{x} cos (x) + e^{x} sin (x) = e^{x} [cos (x) + sin (x)]$

Multiply:

$h^{'} (x) = 2 e^{x} sin (x) \cdot e^{x} [cos (x) + sin (x)] = 2 e^{2 x} sin (x) [cos (x) + sin (x)]$

5. Differentiate

$h (x) = \frac{x}{2 ( x ^{2} + 1 ) ^{2}}$

Solution

(spoiler)

For easier algebra, pull out the constant factor $\frac{1}{2}$ :

$h (x) = \frac{1}{2} \cdot \frac{x}{( x ^{2} + 1 ) ^{2}}$

Now $h (x)$ is a quotient of two functions:

$f (x) = x$
$g (x) = (x^{2} + 1)^{2}$

Use the quotient rule. Note that $g (x)$ is composite, so the chain rule is needed for $g^{'} (x)$ .

Function	Expression
$f (x)$	$x$
$f^{'} (x)$	$1$
$g (x)$	$(x^{2} + 1)^{2}$
$g^{'} (x)$	$2 (x^{2} + 1) \cdot 2 x$

Putting the pieces into the quotient rule:

$h^{'} (x) = \frac{1}{2} \cdot \frac{( x ^{2} + 1 ) ^{2} ( 1 ) - x \cdot 2 ( x ^{2} + 1 ) \cdot 2 x}{( x ^{2} + 1 ) ^{4}}$

$= \frac{( x ^{2} + 1 ) ^{2} - 4 x ^{2} ( x ^{2} + 1 )}{2 ( x ^{2} + 1 ) ^{4}}$

$= \frac{( x ^{2} + 1 ) ( x ^{2} + 1 - 4 x ^{2} )}{2 ( x ^{2} + 1 ) ^{4}}$

$= \frac{1 - 3 x ^{2}}{2 ( x ^{2} + 1 ) ^{3}}$

You can also rewrite a quotient as a product and then use the product rule, if that feels faster. Using the same function as above,

5b. First turn

$h (x) = \frac{x}{2 ( x ^{2} + 1 ) ^{2}}$

into a product of two functions:

$h (x) = \frac{1}{2} x (x^{2} + 1)^{- 2}$

and then find the derivative using the product rule.

Solution

(spoiler)

As before, keep the constant multiple $\frac{1}{2}$ out in front:

$h (x) = \frac{1}{2} x (x^{2} + 1)^{- 2}$

The remaining part is a product of:

$f (x) = x$
$g (x) = (x^{2} + 1)^{- 2}$

Use the product rule. Since $g (x)$ is composite, use the chain rule to find $g^{'} (x)$ .

Function	Expression
$f (x)$	$x$
$f^{'} (x)$	$1$
$g (x)$	$(x^{2} + 1)^{- 2}$
$g^{'} (x)$	$- 2 (x^{2} + 1)^{- 3} \cdot 2 x$

Now apply the product rule:

$h^{'} (x) = \frac{1}{2} \cdot [x (- 4 x (x^{2} + 1)^{- 3}) + 1 (x^{2} + 1)^{- 2}]$

$= \frac{1}{2} \cdot [\frac{- 4 x ^{2}}{( x ^{2} + 1 ) ^{3}} + \frac{1}{( x ^{2} + 1 ) ^{2}}]$

$= \frac{1}{2} \cdot [\frac{- 4 x ^{2}}{( x ^{2} + 1 ) ^{3}} + \frac{( x ^{2} + 1 )}{( x ^{2} + 1 ) ^{3}}]$

$= \frac{1 - 3 x ^{2}}{2 ( x ^{2} + 1 ) ^{3}}$

Some questions ask you to apply derivative rules using only a table of values:

6. $f (x), g (x),$ and their derivatives have the following values shown in the table:

$x$ $f (x)$ $g (x)$ $f^{'} (x)$ $g^{'} (x)$

$1$ $3$ $2$ $- 1$ $5$

$2$ $0.2$ $0.5$ $5$ $3$

$3$ $- 1$ $1$ $4$ $- 3$

Evaluate the first derivative of the function

$h (x) = f (g (x) - x^{2})$

at $x = 1$ .

$x$	$f (x)$	$g (x)$	$f^{'} (x)$	$g^{'} (x)$
$1$	$3$	$2$	$- 1$	$5$
$2$	$0.2$	$0.5$	$5$	$3$
$3$	$- 1$	$1$	$4$	$- 3$

Solution

(spoiler)

Because $h (x) = f (g (x) - x^{2})$ is a composite function, apply the chain rule:

$\frac{d}{d x} [f (g (x) - x^{2})]$

$= f^{'} (g (x) - x^{2}) \cdot (g^{'} (x) - 2 x)$

Now evaluate at $x = 1$ :

$h^{'} (1) = f^{'} (g (1) - 1^{2}) \cdot (g^{'} (1) - 2 (1))$

$= f^{'} (2 - 1) \cdot (5 - 2)$

$= f^{'} (1) \cdot 3$

$= - 1 \cdot 3$

$= - 3$

Challenge problems

1. Differentiate $h (t) = 3 2 π - π t$ .

Solution

(spoiler)

Rewrite the square root as a power:

$h (t) = 3 (2 π - π t)^{1/2}$

The composite function is:

$3 (inner 2 π - π t)^{1/2} outer$

Outer: $f (u) = 3 u^{1/2}$
Inner: $g (t) = 2 π - π t$

Differentiate the outer function:

$f^{'} (u) = \frac{3}{2} u^{- 1/2}$

$f^{'} (g (t)) = \frac{3}{2} (2 π - π t)^{- 1/2}$

Differentiate the inner function:

( $2 π$ is a constant, so its derivative is $0$ . The derivative of $- π t$ is $- π$ )

$g^{'} (t) = - π$

Multiply:

$h^{'} (t) = \frac{3}{2} (2 π - π t)^{- 1/2} (- π)$

$= - \frac{3 π}{2 2 π - π t}$

2. Differentiate $h (x) = \frac{e ^{x^{2}}}{e ^{c o s (x)}}$

Solution

(spoiler)

Instead of using the quotient rule, rewrite the expression using $\frac{a ^{m}}{a ^{n}} = a^{m - n}$ :

$h (x) = e^{(x^{2} - c o s (x))}$

Outer: $f (u) = e^{u}$
Inner: $g (x) = x^{2} - cos (x)$

Differentiate the outer:

$f^{'} (u) = e^{u}$

$f^{'} (g (x)) = e^{x^{2} - c o s (x)}$

Differentiate the inner:

$g^{'} (x) = 2 x + sin (x)$

Multiply:

$h^{'} (x) = e^{(x^{2} - c o s (x))} \cdot (2 x + sin (x))$

This can also be written as

$h^{'} (x) = \frac{e ^{x^{2}} ( 2 x + sin ( x ))}{e ^{c o s (x)}}$

3. Differentiate $A (x) = sin^{3} (ln x)$ .

Solution

(spoiler)

First, rewrite $sin^{3} (ln x)$ as $(sin (ln x))^{3}$ :

$A (x) = (sin (ln x))^{3}$

This function has multiple layers of composition and requires two applications of the chain rule:

Outer: $f (u) = u^{3}$
Middle: $u = sin (v)$
Inner: $v = ln (x)$

Differentiate the outer:

$f^{'} (u) = 3 u^{2}$

Differentiate the middle:

$u^{'} = cos (v)$

Differentiate the inner:

$v^{'} = \frac{1}{x}$

Apply the chain rule layer by layer:

$A^{'} (x) = der. of outer 3 (sin (ln x))^{2} \times cos (ln x) der. of middle \times der. of inner \frac{1}{x}$

$= \frac{3 sin ^{2} ( ln x ) \cdot cos ( ln x )}{x}$