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}.}

Difference between revisions of "Chain Rule"

Latest revision as of 13:20, 15 October 2017

Contents

Introduction

Warm-Up

Exercise 1

Exercise 2

Exercise 3

Exercise 4

Navigation menu

Search

@@ 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==
-Calculate &nbsp;<math style="vertical-align: -5px">f'(x).</math>
+Calculate &nbsp;<math style="vertical-align: -5px">h'(x).</math>
-'''1)''' &nbsp; <math style="vertical-align: -7px">f(x)=(x^2+x+1)(x^3+2x^2+4)</math>
+'''1)''' &nbsp; <math style="vertical-align: -7px">h(x)=\sin(3x)</math>
 {| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
 !Solution: &nbsp;
 |-
-|Using the Product Rule, we have
+|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 style="vertical-align: -5px">f'(x)=\cos(x)</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">g'(x)=3.</math>
-::<math>f'(x)=(x^2+x+1)(x^3+2x^2+4)'+(x^2+x+1)'(x^3+2x^2+4).</math>
-|-
-|Then, using the Power Rule, we have
-|-
-|
-::<math>f'(x)=(x^2+x+1)(3x^2+4x)+(2x+1)(x^3+2x^2+4).</math>
-|-
-|-
-|<u>NOTE:</u> It is not necessary to use the Product Rule to calculate the derivative of this function.
 |-
-|You can distribute the terms and then use the Power Rule.
+|Now, &nbsp;<math style="vertical-align: -6px">h(x)=f\circ g(x).</math>
 |-
-|In this case, we have
+|Using the Chain Rule, we have
 |-
 |
 ::<math>\begin{array}{rcl}
-\displaystyle{f(x)} & = & \displaystyle{(x^2+x+1)(x^3+2x^2+4)}\\
+\displaystyle{h'(x)} & = & \displaystyle{(f'\circ g(x))\cdot g'(x)}\\
-&&\\
-& = & \displaystyle{x^2(x^3+2x^2+4)+x(x^3+2x^2+4)+1(x^3+2x^2+4)}\\
 &&\\
-& = & \displaystyle{x^5+2x^4+4x^2+x^4+2x^3+4x+x^3+2x^2+4} \\
+& = & \displaystyle{\cos (3x)\cdot 3}\\
 &&\\
-& = & \displaystyle{x^5+3x^4+3x^3+6x^2+4x+4.}
+& = & \displaystyle{3\cos (3x).}
 \end{array}</math>
-|-
-|Now, using the Power Rule, we get
-|-
-|
-::<math>f'(x)=5x^4+12x^3+9x^2+12x+4.</math>
-|-
-|In general, calculating derivatives in this way is tedious. It would be better to use the Product Rule.
 |}
@@ Line 71: / Line 53: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math>f'(x)=(x^2+x+1)(3x^2+4x)+(2x+1)(x^3+2x^2+4)</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>h'(x)=3\cos (3x)</math>
-|-
-|or equivalently
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math>f'(x)=x^5+3x^4+3x^3+6x^2+4x+4</math>
 |}
-'''2)''' &nbsp; <math style="vertical-align: -14px">f(x)=\frac{x^2+x^3}{x}</math>
+'''2)''' &nbsp; <math style="vertical-align: -7px">h(x)=(x+1)^8</math>
 {| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
 !Solution: &nbsp;
 |-
-|
+|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>
-Using the Quotient Rule, we have
-|-
-|
-::<math>f'(x)=\frac{x(x^2+x^3)'-(x^2+x^3)(x)'}{x^2}.</math>
-|-
-|Then, using the Power Rule, we have
-|-
-|
-::<math>f'(x)=\frac{x(2x+3x^2)-(x^2+x^3)(1)}{x^2}.</math>
 |-
-|<u>NOTE:</u> It is not necessary to use the Quotient Rule to calculate the derivative of this function.
+|Then, &nbsp;<math style="vertical-align: -5px">f'(x)=8x^7</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">g'(x)=1.</math>
 |-
-|You can divide and then use the Power Rule.
+|Now, &nbsp;<math style="vertical-align: -6px">h(x)=f\circ g(x).</math>
 |-
-|In this case, we have
+|Using the Chain Rule, we have
 |-
 |
 ::<math>\begin{array}{rcl}
-\displaystyle{f(x)} & = & \displaystyle{\frac{x^2+x^3}{x}}\\
+\displaystyle{h'(x)} & = & \displaystyle{(f'\circ g(x))\cdot g'(x)}\\
 &&\\
-& = & \displaystyle{\frac{x^2}{x}+\frac{x^3}{x}}\\
+& = & \displaystyle{8(x+1)^7\cdot 1}\\
 &&\\
-& = & \displaystyle{x+x^2.} \\
+& = & \displaystyle{8(x+1)^7.}
 \end{array}</math>
-|-
-|Now, using the Power Rule, we get
-|-
-|
-::<math>f'(x)=1+2x.</math>
 |}
@@ Line 118: / Line 82: @@
 !Final Answer: &nbsp;
 |-
-||&nbsp; &nbsp; &nbsp; &nbsp;<math>f'(x)=\frac{x(2x+3x^2)-(x^2+x^3)}{x^2}</math>
+||&nbsp; &nbsp; &nbsp; &nbsp;<math>h'(x)=8(x+1)^7.</math>
-|-
-|or equivalently
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math>f'(x)=1+2x</math>
-|-
 |}
-'''3)''' &nbsp; <math style="vertical-align: -14px">f(x)=\frac{\sin x}{\cos x}</math>
+'''3)''' &nbsp; <math style="vertical-align: -7px">h(x)=\ln(x^2)</math>
 {| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
 !Solution: &nbsp;
 |-
-|Using the Quotient Rule, we get
+|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 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 style="vertical-align: -6px">h(x)=f\circ g(x).</math>
+|-
+|Using the Chain Rule, we have
 |-
 |
 ::<math>\begin{array}{rcl}
-\displaystyle{f'(x)} & = & \displaystyle{\frac{\cos x(\sin x)'-\sin x (\cos x)'}{(\cos x)^2}}\\
+\displaystyle{h'(x)} & = & \displaystyle{(f'\circ g(x))\cdot g'(x)}\\
-&&\\
-& = & \displaystyle{\frac{\cos x(\cos x)-\sin x (-\sin x)}{(\cos x)^2}}\\
-&&\\
-& = & \displaystyle{\frac{\cos^2 x+\sin^2 x}{\cos^2 x}} \\
 &&\\
-& = & \displaystyle{\frac{1}{\cos^2 x}}\\
+& = & \displaystyle{\frac{1}{x^2}\cdot 2x}\\
 &&\\
-& = & \displaystyle{\sec^2 x}
+& = & \displaystyle{\frac{2}{x}.}
 \end{array}</math>
-|-
-|since &nbsp;<math style="vertical-align: -2px">\sin^2 x+\cos^2 x=1</math>&nbsp; and &nbsp;<math style="vertical-align: -13px">\sec x=\frac{1}{\cos x}.</math>
-|-
-|Since &nbsp;<math style="vertical-align: -14px">\frac{\sin x}{\cos x}=\tan x,</math>&nbsp; we have
-|-
-|
-::<math>\frac{d}{dx}{\tan x}=\sec^2 x.</math>
 |}
@@ Line 158: / Line 111: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math>f'(x)=\sec^2 x</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>h'(x)=\frac{2}{x}</math>
 |-
 |}
@@ Line 164: / 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
-::<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{\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{-\csc x \cot x.}
-\end{array}</math>
-Using the Product Rule and Power Rule, we have
+Using the Chain 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{h'(x)} & = & \displaystyle{4(\sin x+\cos x)^3 (\sin x+\cos x)'}\\
 &&\\
-& = & \displaystyle{\frac{1}{x^2}(-\csc x \cot x+0)+(-2x^{-3})(\csc x-4)}\\
+& = & \displaystyle{4(\sin x+\cos x)^3 ((\sin x)'+(\cos x)')}\\
 &&\\
-& = & \displaystyle{\frac{-\csc x \cot x}{x^2}+\frac{-2(\csc x-4)}{x^3}.}
+& = & \displaystyle{4(\sin x+\cos x)^3 (\cos x-\sin x)}.
 \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.
-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>
+First, notice &nbsp;<math style="vertical-align: -6px">h(x)=(\sin(2x^2+x+1))^3.</math>
-Using the Product Rule, we get
+Using the Chain Rule, we have
-::<math>\begin{array}{rcl}
+::<math>h'(x)=3(\sin(2x^2+x+1))^2 \cdot (\sin(2x^2+x+1))'.</math>
-\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,
+Now, we need to use the Chain Rule a second time. So, we get
 ::<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>