MathAdmin: Created page with "==Introduction== It is relatively easy to calculate the derivatives of simple functions, like polynomials or trigonometric functions. But, what about more complicat..."

2017-10-15T21:44:38Z

Created page with "==Introduction== It is relatively easy to calculate the derivatives of <em>simple functions</em>, like polynomials or trigonometric functions. But, what about more complicat..."

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==Introduction==
It is relatively easy to calculate the derivatives of <em>simple functions</em>, like polynomials or trigonometric functions.

But, what about more complicated functions?

For example,  <math style="vertical-align: -5px">f(x)=\sin(3x)</math>  or  <math style="vertical-align: -5px">g(x)=(x+1)^8?</math>

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

For  <math style="vertical-align: -5px">f(x)=\sin(3x),</math>  it is the composition of the function  <math style="vertical-align: -4px">y=3x</math>  with  <math style="vertical-align: -5px">y=\sin(x).</math>

Similarly, for  <math style="vertical-align: -5px">g(x)=(x+1)^8,</math>  it is the composition of  <math style="vertical-align: -5px">y=x+1</math>  and  <math style="vertical-align: -5px">y=x^8.</math>

So, how do we take the derivative of compositions?

The answer to this question is exactly the Chain Rule.

'''Chain Rule'''

Let  <math style="vertical-align: -6px">y=f(u)</math>  be a differentiable function of  <math style="vertical-align: -1px">u</math>  and let  <math style="vertical-align: -6px">u=g(x)</math>  be a differentiable function of  <math style="vertical-align: -1px">x.</math> 

Then,  <math style="vertical-align: -5px">y=f\circ g(x))</math>  is a differentiable function of  <math style="vertical-align: 0px">x</math>  and

::<math>y'=(f'\circ g(x))\cdot g'(x).</math>

==Warm-Up==
Calculate  <math style="vertical-align: -5px">h'(x).</math>

'''1)'''   <math style="vertical-align: -7px">h(x)=\sin(3x)</math>

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Solution:  
|-
|Let  <math style="vertical-align: -5px">f(x)=\sin (x)</math>  and  <math style="vertical-align: -5px">g(x)=3x.</math>
|-
|Then,  <math style="vertical-align: -5px">f'(x)=\cos(x)</math>  and  <math style="vertical-align: -5px">g'(x)=3.</math>
|-
|Now,  <math style="vertical-align: -6px">h(x)=f\circ g(x).</math>
|-
|Using the Chain Rule, we have
|-
|
::<math>\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}</math>
|}

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|       <math>h'(x)=3\cos (3x)</math>
|}

'''2)'''   <math style="vertical-align: -7px">h(x)=(x+1)^8</math>

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Solution:  
|-
|Let  <math style="vertical-align: -5px">f(x)=x^8</math>  and  <math style="vertical-align: -5px">g(x)=x+1.</math>
|-
|Then,  <math style="vertical-align: -5px">f'(x)=8x^7</math>  and  <math style="vertical-align: -5px">g'(x)=1.</math>
|-
|Now,  <math style="vertical-align: -6px">h(x)=f\circ g(x).</math>
|-
|Using the Chain Rule, we have
|-
|
::<math>\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}</math>
|}

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
||       <math>h'(x)=8(x+1)^7.</math>
|}

'''3)'''   <math style="vertical-align: -7px">h(x)=\ln(x^2)</math>

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Solution:  
|-
|Let  <math style="vertical-align: -5px">f(x)=\ln (x)</math>  and  <math style="vertical-align: -5px">g(x)=x^2.</math>
|-
|Then,  <math style="vertical-align: -13px">f'(x)=\frac{1}{x}</math>  and  <math style="vertical-align: -5px">g'(x)=2x.</math>
|-
|Now,  <math style="vertical-align: -6px">h(x)=f\circ g(x).</math>
|-
|Using the Chain Rule, we have
|-
|
::<math>\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}</math>
|}

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|       <math>h'(x)=\frac{2}{x}</math>
|-
|}

== Exercise 1 ==

Calculate the derivative of  <math style="vertical-align: -6px">h(x)=(\sin x+\cos x)^4.</math>

Using the Chain Rule, we have

::<math>\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}</math>

So, we have
::<math>h'(x)=4(\sin x+\cos x)^3 (\cos x-\sin x).</math>

== Exercise 2 ==

Calculate the derivative of  <math style="vertical-align: -6px">h(x)=\sin^3(2x^2+x+1).</math>

First, notice  <math style="vertical-align: -6px">h(x)=(\sin(2x^2+x+1))^3.</math>

Using the Chain Rule, we have

::<math>h'(x)=3(\sin(2x^2+x+1))^2 \cdot (\sin(2x^2+x+1))'.</math>

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

::<math>\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}</math>

So, we have
::<math>h'(x)=3\sin^2(2x^2+x+1) \cos(2x^2+x+1)(4x+1).</math>

== Exercise 3 ==

Calculate the derivative of  <math style="vertical-align: -6px">h(x)=\cos (2x+1)\sin(x^2+3x).</math>

Using the Product Rule, we have

::<math>h'(x)=\cos(2x+1)(\sin(x^2+3x))'+(\cos(2x+1))'\sin(x^2+3x).</math>

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

So, using the Chain Rule, we have

::<math>\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}</math>

So, we get
::<math>h'(x)=\cos(2x+1)\cos(x^2+3x) (2x+3)-\sin(2x+1)(2)\sin(x^2+3x).</math>

== Exercise 4 ==

Calculate the derivative of  <math style="vertical-align: -16px">h(x)=\frac{\sin(3x)+x\cos(2x)}{x^2+1}.</math>

First, using the Quotient Rule, we have

::<math>\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}</math>

Using the Product Rule, we get

::<math>\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}</math>

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

::<math>\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}</math>

So, we have
::<math>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}.</math>

Chain Rule - Revision history

MathAdmin: Created page with "==Introduction== It is relatively easy to calculate the derivatives of simple functions, like polynomials or trigonometric functions. But, what about more complicat..."