Chain Rule

Introduction

It is relatively easy to calculate the derivatives of simple functions, like polynomials or trigonometric functions.

But, what about more complicated functions?

For example,  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f(x)=\sin(3x)}   or  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle g(x)=(x+1)^{8}?}

Well, the key to calculating the derivatives of these functions is to recognize that these functions are compositions.

For  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f(x)=\sin(3x),}   it is the composition of the function  $y=3x$  with  $y=\sin(x).$

Similarly, for  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle g(x)=(x+1)^{8},}   it is the composition of  $y=x+1$  and  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle y=x^{8}.}

So, how do we take the derivative of compositions?

The answer to this question is exactly the Chain Rule.

Chain Rule

Let  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle y=f(u)}   be a differentiable function of  $u$  and let  $u=g(x)$  be a differentiable function of  $x.$ 

Then,  $y=f\circ g(x))$  is a differentiable function of  $x$  and

$y'=(f'\circ g(x))\cdot g'(x).$

Warm-Up

Calculate  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle h'(x).}

1)   Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle h(x)=\sin(3x)}

Solution:
Let  $f(x)=\sin(x)$  and  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle g(x)=3x.}
Then,  $f'(x)=\cos(x)$  and  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle g'(x)=3.}
Now,  $h(x)=f\circ g(x).$
Using the Chain Rule, we have
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}\displaystyle {h'(x)}&=&\displaystyle {(f'\circ g(x))\cdot g'(x)}\\&&\\&=&\displaystyle {\cos(3x)\cdot 3}\\&&\\&=&\displaystyle {3\cos(3x).}\end{array}}}

Final Answer:
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle h'(x)=3\cos(3x)}

2)   Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle h(x)=(x+1)^{8}}

Solution:
Let  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f(x)=x^{8}}   and  $g(x)=x+1.$
Then,  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f'(x)=8x^{7}}   and  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle g'(x)=1.}
Now,  $h(x)=f\circ g(x).$
Using the Chain Rule, we have
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}\displaystyle {h'(x)}&=&\displaystyle {(f'\circ g(x))\cdot g'(x)}\\&&\\&=&\displaystyle {8(x+1)^{7}\cdot 1}\\&&\\&=&\displaystyle {8(x+1)^{7}.}\end{array}}}

Final Answer:
$h'(x)=8(x+1)^{7}.$

3)   Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle h(x)=\ln(x^{2})}

Solution:
Let  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f(x)=\ln(x)}   and  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g(x)=x^2.}
Then,  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'(x)=\frac{1}{x}}   and  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g'(x)=2x.}
Now,  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h(x)=f\circ g(x).}
Using the Chain Rule, we have
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{h'(x)} & = & \displaystyle{(f'\circ g(x))\cdot g'(x)}\\ &&\\ & = & \displaystyle{\frac{1}{x^2}\cdot 2x}\\ &&\\ & = & \displaystyle{\frac{2}{x}.} \end{array}}

Final Answer:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h'(x)=\frac{2}{x}}

Exercise 1

Calculate the derivative of  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h(x)=(\sin x+\cos x)^4.}

Using the Chain Rule, we have

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{h'(x)} & = & \displaystyle{4(\sin x+\cos x)^3 (\sin x+\cos x)'}\\ &&\\ & = & \displaystyle{4(\sin x+\cos x)^3 ((\sin x)'+(\cos x)')}\\ &&\\ & = & \displaystyle{4(\sin x+\cos x)^3 (\cos x-\sin x)}. \end{array}}

So, we have

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h'(x)=4(\sin x+\cos x)^3 (\cos x-\sin x).}

Exercise 2

Calculate the derivative of  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h(x)=\sin^3(2x^2+x+1).}

First, notice  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h(x)=(\sin(2x^2+x+1))^3.}

Using the Chain Rule, we have

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h'(x)=3(\sin(2x^2+x+1))^2 \cdot (\sin(2x^2+x+1))'.}

Now, we need to use the Chain Rule a second time. So, we get

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{h'(x)} & = & \displaystyle{3(\sin(2x^2+x+1))^2 \cos(2x^2+x+1)\cdot (2x^2+x+1)'}\\ &&\\ & = & \displaystyle{3\sin^2(2x^2+x+1) \cos(2x^2+x+1)(4x+1).} \end{array}}

So, we have

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h'(x)=3\sin^2(2x^2+x+1) \cos(2x^2+x+1)(4x+1).}

Exercise 3

Calculate the derivative of  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h(x)=\cos (2x+1)\sin(x^2+3x).}

Using the Product Rule, we have

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h'(x)=\cos(2x+1)(\sin(x^2+3x))'+(\cos(2x+1))'\sin(x^2+3x).}

For the two remaining derivatives, we need to use the Chain Rule.

So, using the Chain Rule, we have

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{h'(x)} & = & \displaystyle{\cos(2x+1)\cos(x^2+3x)\cdot (x^2+3x)'-\sin(2x+1)\cdot (2x+1)'\sin(x^2+3x)}\\ &&\\ & = & \displaystyle{\cos(2x+1)\cos(x^2+3x) (2x+3)-\sin(2x+1)(2)\sin(x^2+3x).} \end{array}}

So, we get

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h'(x)=\cos(2x+1)\cos(x^2+3x) (2x+3)-\sin(2x+1)(2)\sin(x^2+3x).}

Exercise 4

Calculate the derivative of  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h(x)=\frac{\sin(3x)+x\cos(2x)}{x^2+1}.}

First, using the Quotient Rule, we have

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{h'(x)} & = & \displaystyle{\frac{(x^2+1)(\sin(3x)+x\cos(2x))'-(\sin(3x)+x\cos(2x))(x^2+1)'}{(x^2+1)^2}}\\ &&\\ & = & \displaystyle{\frac{(x^2+1)[(\sin(3x))'+(x\cos(2x))']-(\sin(3x)+x\cos(2x))(2x)}{(x^2+1)^2}.} \end{array}}

Using the Product Rule, we get

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{h'(x)} & = & \displaystyle{\frac{(x^2+1)[(\sin(3x))'+x(\cos(2x))'+(x)'\cos(2x)]-(\sin(3x)+x\cos(2x))(2x)}{(x^2+1)^2}}\\ &&\\ & = & \displaystyle{\frac{(x^2+1)[(\sin(3x))'+x(\cos(2x))'+1\cos(2x)]-(\sin(3x)+x\cos(2x))(2x)}{(x^2+1)^2}.} \end{array}}

For the remaining derivatives, we need to use the Chain Rule. So, we get

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{h'(x)} & = & \displaystyle{\frac{(x^2+1)[\cos(3x)(3x)'+x(-\sin(2x))(2x)'+\cos(2x)]-(\sin(3x)+x\cos(2x))(2x)}{(x^2+1)^2}}\\ &&\\ & = & \displaystyle{\frac{(x^2+1)[\cos(3x)(3)-x\sin(2x)(2)+\cos(2x)]-(\sin(3x)+x\cos(2x))(2x)}{(x^2+1)^2}.} \end{array}}

So, we have

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h'(x)=\frac{(x^2+1)[\cos(3x)(3)-x\sin(2x)(2)+\cos(2x)]-(\sin(3x)+x\cos(2x))(2x)}{(x^2+1)^2}.}