Difference between revisions of "Chain Rule"

Latest revision as of 13:20, 15 October 2017

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, $f(x)=\sin(3x)$ or 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+1)^8?}

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

For 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)=\sin(3x),} it is the composition of the function 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 y=3x} with 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 y=\sin(x).}

Similarly, for 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+1)^8,} it is the composition 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 y=x+1} 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 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 (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 y=f(u)} be a differentiable function 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 u} and let 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 u=g(x)} be a differentiable function 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 x.}

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 y=f\circ g(x))} is a differentiable function 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 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 y'=(f'\circ g(x))\cdot g'(x).}

Warm-Up

Calculate 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).}

1) 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(3x)}

Solution:
Let 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)=\sin (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)=3x.}
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)=\cos(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)=3.}
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{\cos (3x)\cdot 3}\\ &&\\ & = & \displaystyle{3\cos (3x).} \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)=3\cos (3x)}

2) 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)=(x+1)^8}

Solution:
Let 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)=x^8} 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+1.}
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)=8x^7} 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)=1.}
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{8(x+1)^7\cdot 1}\\ &&\\ & = & \displaystyle{8(x+1)^7.} \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)=8(x+1)^7.}

3) 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)=\ln(x^2)}

Solution:
Let 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)=\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}.}

@@ Line 20: / Line 20: @@
 Let &nbsp;<math style="vertical-align: -6px">y=f(u)</math>&nbsp; be a differentiable function of &nbsp;<math style="vertical-align: -1px">u</math>&nbsp; and let &nbsp;<math style="vertical-align: -6px">u=g(x)</math>&nbsp; be a differentiable function of &nbsp;<math style="vertical-align: -1px">x.</math>&nbsp;
-Then, &nbsp;<math style="vertical-align: -5px">y=f(g(x))</math>&nbsp; is a differentiable function of &nbsp;<math style="vertical-align: 0px">x</math>&nbsp; and
+Then, &nbsp;<math style="vertical-align: -5px">y=f\circ g(x))</math>&nbsp; is a differentiable function of &nbsp;<math style="vertical-align: 0px">x</math>&nbsp; and
-::<math>y'=f'(g(x))\cdot g'(x).</math>
+::<math>y'=(f'\circ g(x))\cdot g'(x).</math>
 ==Warm-Up==
@@ Line 32: / Line 32: @@
 !Solution: &nbsp;
 |-
-|Let &nbsp;<math>f(x)=\sin (x)</math>&nbsp; and &nbsp;<math>g(x)=3x.</math>
+|Let &nbsp;<math style="vertical-align: -5px">f(x)=\sin (x)</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">g(x)=3x.</math>
 |-
-|Then, &nbsp;<math>f'(x)=\cos(x)</math>&nbsp; and &nbsp;<math>g'(x)=3.</math>
+|Then, &nbsp;<math style="vertical-align: -5px">f'(x)=\cos(x)</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">g'(x)=3.</math>
 |-
-|Now, &nbsp;<math>h(x)=f(g(x)).</math>
+|Now, &nbsp;<math style="vertical-align: -6px">h(x)=f\circ g(x).</math>
 |-
 |Using the Chain Rule, we have
@@ Line 42: / Line 42: @@
 |
 ::<math>\begin{array}{rcl}
-\displaystyle{h'(x)} & = & \displaystyle{f'(g(x))\cdot g'(x)}\\
+\displaystyle{h'(x)} & = & \displaystyle{(f'\circ g(x))\cdot g'(x)}\\
 &&\\
 & = & \displaystyle{\cos (3x)\cdot 3}\\
@@ Line 61: / Line 61: @@
 !Solution: &nbsp;
 |-
-|Let &nbsp;<math>f(x)=x^8</math>&nbsp; and &nbsp;<math>g(x)=x+1.</math>
+|Let &nbsp;<math style="vertical-align: -5px">f(x)=x^8</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">g(x)=x+1.</math>
 |-
-|Then, &nbsp;<math>f'(x)=8x^7</math>&nbsp; and &nbsp;<math>g'(x)=1.</math>
+|Then, &nbsp;<math style="vertical-align: -5px">f'(x)=8x^7</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">g'(x)=1.</math>
 |-
-|Now, &nbsp;<math>h(x)=f(g(x)).</math>
+|Now, &nbsp;<math style="vertical-align: -6px">h(x)=f\circ g(x).</math>
 |-
 |Using the Chain Rule, we have
@@ Line 71: / Line 71: @@
 |
 ::<math>\begin{array}{rcl}
-\displaystyle{h'(x)} & = & \displaystyle{f'(g(x))\cdot g'(x)}\\
+\displaystyle{h'(x)} & = & \displaystyle{(f'\circ g(x))\cdot g'(x)}\\
 &&\\
 & = & \displaystyle{8(x+1)^7\cdot 1}\\
@@ Line 90: / Line 90: @@
 !Solution: &nbsp;
 |-
-|Let &nbsp;<math>f(x)=\ln (x)</math>&nbsp; and &nbsp;<math>g(x)=x^2.</math>
+|Let &nbsp;<math style="vertical-align: -5px">f(x)=\ln (x)</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">g(x)=x^2.</math>
 |-
-|Then, &nbsp;<math>f'(x)=\frac{1}{x}</math>&nbsp; and &nbsp;<math>g'(x)=2x.</math>
+|Then, &nbsp;<math style="vertical-align: -13px">f'(x)=\frac{1}{x}</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">g'(x)=2x.</math>
 |-
-|Now, &nbsp;<math>h(x)=f(g(x)).</math>
+|Now, &nbsp;<math style="vertical-align: -6px">h(x)=f\circ g(x).</math>
 |-
 |Using the Chain Rule, we have
@@ Line 100: / Line 100: @@
 |
 ::<math>\begin{array}{rcl}
-\displaystyle{h'(x)} & = & \displaystyle{f'(g(x))\cdot g'(x)}\\
+\displaystyle{h'(x)} & = & \displaystyle{(f'\circ g(x))\cdot g'(x)}\\
 &&\\
 & = & \displaystyle{\frac{1}{x^2}\cdot 2x}\\
@@ Line 117: / Line 117: @@
 == Exercise 1 ==
-Calculate the derivative of &nbsp;<math style="vertical-align: -13px">f(x)=\frac{1}{x^2}(\csc x-4).</math>
+Calculate the derivative of &nbsp;<math style="vertical-align: -6px">h(x)=(\sin x+\cos x)^4.</math>
-First, we need to know the derivative of &nbsp;<math style="vertical-align: 0px">\csc x.</math>&nbsp; Recall
+Using the Chain Rule, we have
-::<math>\csc x =\frac{1}{\sin x}.</math>
-Now, using the Quotient Rule, we have
 ::<math>\begin{array}{rcl}
-\displaystyle{\frac{d}{dx}(\csc x)} & = & \displaystyle{\frac{d}{dx}\bigg(\frac{1}{\sin x}\bigg)}\\
+\displaystyle{h'(x)} & = & \displaystyle{4(\sin x+\cos x)^3 (\sin x+\cos x)'}\\
-&&\\
-& = & \displaystyle{\frac{\sin x (1)'-1(\sin x)'}{\sin^2 x}}\\
-&&\\
-& = & \displaystyle{\frac{\sin x (0)-\cos x}{\sin^2 x}}\\
 &&\\
-& = & \displaystyle{\frac{-\cos x}{\sin^2 x}} \\
+& = & \displaystyle{4(\sin x+\cos x)^3 ((\sin x)'+(\cos x)')}\\
 &&\\
-& = & \displaystyle{-\csc x \cot x.}
+& = & \displaystyle{4(\sin x+\cos x)^3 (\cos x-\sin x)}.
-\end{array}</math>
-Using the Product Rule and Power Rule, we have
-::<math>\begin{array}{rcl}
-\displaystyle{f'(x)} & = & \displaystyle{\frac{1}{x^2}(\csc x-4)'+\bigg(\frac{1}{x^2}\bigg)'(\csc x-4)}\\
-&&\\
-& = & \displaystyle{\frac{1}{x^2}(-\csc x \cot x+0)+(-2x^{-3})(\csc x-4)}\\
-&&\\
-& = & \displaystyle{\frac{-\csc x \cot x}{x^2}+\frac{-2(\csc x-4)}{x^3}.}
 \end{array}</math>
 So, we have
-::<math>f'(x)=\frac{-\csc x \cot x}{x^2}+\frac{-2(\csc x-4)}{x^3}.</math>
+::<math>h'(x)=4(\sin x+\cos x)^3 (\cos x-\sin x).</math>
 == Exercise 2 ==
-Calculate the derivative of &nbsp;<math style="vertical-align: -5px">g(x)=2x\sin x \sec x.</math>
+Calculate the derivative of &nbsp;<math style="vertical-align: -6px">h(x)=\sin^3(2x^2+x+1).</math>
-Notice that the function &nbsp;<math style="vertical-align: -5px">g(x)</math>&nbsp; is the product of three functions.
+First, notice &nbsp;<math style="vertical-align: -6px">h(x)=(\sin(2x^2+x+1))^3.</math>
-We start by grouping two of the functions together. So, we have &nbsp;<math style="vertical-align: -5px">g(x)=(2x\sin x)\sec x.</math>
+Using the Chain Rule, we have
-Using the Product Rule, we get
+::<math>h'(x)=3(\sin(2x^2+x+1))^2 \cdot (\sin(2x^2+x+1))'.</math>
-::<math>\begin{array}{rcl}
+Now, we need to use the Chain Rule a second time. So, we get
-\displaystyle{g'(x)} & = & \displaystyle{(2x\sin x)(\sec x)'+(2x\sin x)'\sec x}\\
-&&\\
-& = & \displaystyle{(2x\sin x)(\tan^2 x)+(2x\sin x)'\sec x.}
-\end{array}</math>
-Now, we need to use the Product Rule again. So,
 ::<math>\begin{array}{rcl}
-\displaystyle{g'(x)} & = & \displaystyle{2x\sin x\tan^2 x+(2x(\sin x)'+(2x)'\sin x)\sec x}\\
+\displaystyle{h'(x)} & = & \displaystyle{3(\sin(2x^2+x+1))^2 \cos(2x^2+x+1)\cdot (2x^2+x+1)'}\\
 &&\\
-& = & \displaystyle{2x\sin x\tan^2 x+(2x\cos x+2\sin x)\sec x.}
+& = & \displaystyle{3\sin^2(2x^2+x+1) \cos(2x^2+x+1)(4x+1).}
 \end{array}</math>
 So, we have
-::<math>g'(x)=2x\sin x\tan^2 x+(2x\cos x+2\sin x)\sec x.</math>
+::<math>h'(x)=3\sin^2(2x^2+x+1) \cos(2x^2+x+1)(4x+1).</math>
-But, there is another way to do this problem. Notice
+== Exercise 3 ==
-::<math>\begin{array}{rcl}
-\displaystyle{g(x)} & = & \displaystyle{2x\sin x\sec x}\\
-&&\\
-& = & \displaystyle{2x\sin x\frac{1}{\cos x}}\\
-&&\\
-& = & \displaystyle{2x\tan x.}
-\end{array}</math>
-Now, you would only need to use the Product Rule once instead of twice.
-== Exercise 3 ==
+Calculate the derivative of &nbsp;<math style="vertical-align: -6px">h(x)=\cos (2x+1)\sin(x^2+3x).</math>
-Calculate the derivative of &nbsp;<math style="vertical-align: -16px">h(x)=\frac{x^2\sin x+1}{x^2\cos x+3}.</math>
+Using the Product Rule, we have
-Using the Quotient Rule, we have
+::<math>h'(x)=\cos(2x+1)(\sin(x^2+3x))'+(\cos(2x+1))'\sin(x^2+3x).</math>
-::<math>h'(x)=\frac{(x^2\cos x+3)(x^2\sin x+1)'-(x^2\sin x+1)(x^2\cos x+3)'}{(x^2\cos x+3)^2}.</math>
+For the two remaining derivatives, we need to use the Chain Rule.
-Now, we need to use the Product Rule. So, we have
+So, using the Chain Rule, we have
 ::<math>\begin{array}{rcl}
-\displaystyle{h'(x)} & = & \displaystyle{\frac{(x^2\cos x+3)(x^2(\sin x)'+(x^2)'\sin x)-(x^2\sin x+1)(x^2(\cos x)'+(x^2)'\cos x)}{(x^2\cos x+3)^2}}\\
+\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{\frac{(x^2\cos x+3)(x^2\cos x+2x\sin x)-(x^2\sin x+1)(-x^2\sin x+2x\cos x)}{(x^2\cos x+3)^2}.}
+& = & \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)=\frac{(x^2\cos x+3)(x^2\cos x+2x\sin x)-(x^2\sin x+1)(-x^2\sin x+2x\cos x)}{(x^2\cos x+3)^2}.</math>
+::<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  &nbsp;<math style="vertical-align: -14px">f(x)=\frac{e^x}{x^2\sin x}.</math>
+Calculate the derivative of  &nbsp;<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{f'(x)} & = & \displaystyle{\frac{x^2\sin x (e^x)'-e^x(x^2\sin x)'}{(x^2\sin x)^2}}\\
+\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\sin x e^x - e^x(x^2\sin x)'}{x^4\sin^2 x}.}
+& = & \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>
-Now, we need to use the Product Rule. So, we have
+For the remaining derivatives, we need to use the Chain Rule. So, we get
 ::<math>\begin{array}{rcl}
-\displaystyle{f'(x)} & = & \displaystyle{\frac{x^2\sin x e^x - e^x(x^2(\sin x)'+(x^2)'\sin x)}{x^4\sin^2 x}}\\
+\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\sin x e^x - e^x(x^2\cos x+2x\sin x)}{x^4\sin^2 x}.}
+& = & \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>f'(x)=\frac{x^2\sin x e^x - e^x(x^2\cos x+2x\sin x)}{x^4\sin^2 x}.</math>
+::<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>

Difference between revisions of "Chain Rule"

Latest revision as of 13:20, 15 October 2017

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Warm-Up

Exercise 1

Exercise 2

Exercise 3

Exercise 4

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