Chain rule

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

$$\underset{\text{outer }f}{\underbrace{(\stackrel{\text{inner }g}{\overbrace{\boxed{2x+1}}}{)}^3}}$$

For clarity,

Outer: $f(g)=g^3$
Inner: $g(x)=2x+1$

1. Differentiate the outer function with the power rule:

$$f^{\prime }(g)=3g^2$$

The inner function should remain unchanged, so replace the placeholder letter $g$ with the function $g(x)$.

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

2. Differentiate the inner function:

$$g^{\prime }(x)=2$$

3. Multiply together:

$$\begin{array}{rl}{h}^{\prime }(x)& =f^{\prime }(g(x))\cdot g^{\prime }(x)\\ & =3(2x+1{)}^2\cdot 2\\ & =\boxed{6(2x+1{)}^2}\end{array}$$

$h(x)$ can be written as a composite function $f(g(x))$ with:

$$\underset{\text{outer }f}{\underbrace{\ln (\stackrel{\text{inner }g}{\overbrace{\boxed{3x^2+x}}})}}$$

1. Differentiate the outer function, keeping the inner one unchanged:

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

2. Differentiate the inner function:

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

3. Multiply:

$$\boxed{h^{\prime }(x)=\frac{6x+1}{3x^2+x}}$$

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

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

a) For $a(x)$, the outer function is $f(g(x))=e^{g(x)}$ with the inner function $g(x)=-2x$.

1. Differentiate the outer:

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

2. Differentiate the inner:

$$g^{\prime }(x)=-2$$

3. Multiply:

$$\begin{array}{rl}{h}^{\prime }(x)& =f^{\prime }(g(x))\cdot g^{\prime }(x)\\ & =\boxed{-2e^{-2x}}\end{array}$$

b) For $b(x)$, the outer function is $f(g(x))=2^{g(x)}$ with the inner function $g(x)=3x+1$.

1. Differentiate the outer:

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

2. Differentiate the inner:

$$g^{\prime }(x)=3$$

3. Multiply:

$$\begin{array}{rl}{b}^{\prime }(x)& =f^{\prime }(g(x))\cdot g^{\prime }(x)\\ & =(2^{3x+1}\cdot \ln (2))\cdot (3)\\ & =\boxed{(3\ln (2))\cdot 2^{3x+1}}\end{array}$$

$h(x)$ is primarily a product of the two functions $f(x)=3x$ and $g(x)=(x^2-1{)}^3$. Note that $g(x)$ is a composite function so the chain rule must be used when taking its derivative.

Putting all of the pieces into the product rule:

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

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

Just to simplify further, notice both terms share common factors $3$ and $(x^2-1{)}^2$. Factor those out:

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

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

$h(x)$ is a composite function with

$$\underset{\text{outer }f}{\underbrace{\left(\stackrel{\text{inner }g}{\overbrace{\boxed{e^x\sin (x)}}}\right){}^2}}$$

1. Differentiate outer:

$$f^{\prime }(g(x))=2(\underset{g(x)}{\underbrace{e^x\sin (x)}}{)}^1$$

2. Differentiate inner:

$g(x)$ is a product of the two functions $e^x$ and $\sin (x)$. Applying the product rule,

$$\begin{gathered} g^{\prime }(x)=e^x\cos (x)+e^x\sin (x) \\ =e^x[\cos (x)+\sin (x)] \end{gathered}$$

3. Multiply:

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

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

For ease of calculation, let’s bring the $\frac{1}{2}$ out as a constant multiple.

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

$h(x)$ is primarily a quotient of two functions $f(x)=x$ and $g(x)=(x^2+1{)}^2$, so use the quotient rule first. Here are the pieces. Note that $g(x)$ is a composite function so the chain rule must be used when taking its derivative.

Putting the pieces into the quotient rule:

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

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

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

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

Just like before, take out the constant multiple $\frac{1}{2}$ and leave it in front for ease of calculation. The remaining portion of $h(x)$ is the product of $f(x)=x$ and $g(x)=(x^2+1{)}^{-2}$ so use the product rule first. Again, $g(x)$ is a composite function so use the chain rule. Here are the pieces:

Into the product rule they go:

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

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

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

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

Answer: -3

Because $f$ is a composite function, the chain rule is applied when taking the derivative:

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

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

Evaluating at $x=1$:

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

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

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

$$=-1\cdot 3$$

$$=\boxed{-3}$$

$h(t)$ can be rewritten as $3(2\pi -\pi t{)}^{1/2}$ with:

$$\stackrel{\text{outer}}{\overbrace{3{(\underset{\text{inner}}{\underbrace{\boxed{2\pi -\pi t}}})}^{1/2}}}$$

For clarity,

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

1. Differentiate outer:

$$f^{\prime }(g(t))=\frac{3}{2}(2\pi -\pi t{)}^{-1/2}$$

2. Differentiate inner:

$$g^{\prime }(t)=-\pi$$

Note: Remember that $2\pi$ is a constant so the derivative is $0$, but $-\pi$ is the coefficient in front of the linear term with variable $t$.

3. Multiply:

$$h^{\prime }(t)=\frac{3}{2}(2\pi -\pi t{)}^{-1/2}(-\pi )$$

$$=\boxed{-\frac{3\pi }{2\sqrt{2\pi -\pi t}}}$$

Rather than using the quotient rule, it would be more efficient to use the exponent rule $\frac{a^m}{a^n}=a^{m-n}$ to rewrite the function into

$$h(x)=e^{(x^2-\cos (x))}$$

Then applying the chain rule,

$$h^{\prime }(x)=e^{(x^2-\cos (x))}\cdot (2x+\sin (x))$$

Which can also be written as

$$\boxed{h^{\prime }(x)=\frac{e^{x^2}(2x+\sin (x))}{e^{\cos (x)}}}$$

First, ${\sin }^3(\ln x)$ is the same thing as $(\sin (\ln x){)}^3$.

$A(x)=(\sin (\ln x){)}^3$ contains several layers of composition:

Outer: $f(g)=g^3$
Middle: $g(h)=\sin (h)$
Inner: $h(x)=\ln (x)$

So the chain rule will be applied multiple times.

Let’s peel away the layers of this composite function with the chain rule:

$$A^{\prime }(x)=\underset{\text{der. of outer }f}{\underbrace{3(\sin (\ln x){)}^2}}\times \stackrel{\text{der. of middle }g}{\overbrace{\cos (\ln x)}}\times \underset{\text{der. of inner }h}{\underbrace{\frac{1}{x}}}$$

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