Chain Rule - Revision history

Kayla Murray at 21:20, 15 October 2017

2017-10-15T21:20:02Z

Kayla Murray: /* Exercise 4 */

2017-10-09T17:31:00Z

Exercise 4

Kayla Murray: /* Exercise 3 */

2017-10-09T17:20:50Z

Exercise 3

Kayla Murray: /* Exercise 2 */

2017-10-09T17:14:54Z

Exercise 2

Kayla Murray: /* Exercise 1 */

2017-10-09T17:07:15Z

Exercise 1

Kayla Murray: /* Warm-Up */

2017-10-09T17:02:34Z

Warm-Up

Kayla Murray: /* Warm-Up */

2017-10-09T16:47:42Z

Warm-Up

Kayla Murray: /* Warm-Up */

2017-10-09T16:24:30Z

Warm-Up

Kayla Murray: /* Introduction */

2017-10-09T16:21:25Z

Introduction

Kayla Murray: /* Introduction */

2017-10-09T16:14:29Z

Introduction

@@ 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 36: / Line 36: @@
 |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 style="vertical-align: -6px">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 65: / Line 65: @@
 |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 style="vertical-align: -6px">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 94: / Line 94: @@
 |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(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 176: / Line 176: @@
 == 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\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))']-(\sin(3x)+x\cos(2x))(2x)}{(x^2+1)^2}.}
 \end{array}</math>
-Now, we need to use the Product Rule. So, we have
+Using the Product Rule, 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)[(\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\cos x+2x\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>
 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>

@@ Line 155: / Line 155: @@
 == Exercise 3 ==
-Calculate the derivative of &nbsp;<math style="vertical-align: -16px">h(x)=\frac{x^2\sin x+1}{x^2\cos x+3}.</math>
+Calculate the derivative of &nbsp;<math style="vertical-align: -6px">h(x)=\cos (2x+1)\sin(x^2+3x).</math>
-Using the Quotient Rule, we have
+Using the Product Rule, we have
-::<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>
+::<math>h'(x)=\cos(2x+1)(\sin(x^2+3x))'+(\cos(2x+1))'\sin(x^2+3x).</math>
-Now, we need to use the Product Rule. So, we have
+For the two remaining derivatives, we need to use the Chain Rule.
 ::<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 ==

@@ Line 134: / Line 134: @@
 == 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
 ::<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>
 == Exercise 3 ==

@@ 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>\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 ==

@@ 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(g(x)).</math>
 |-
 |Using the Chain Rule, we have
@@ 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(g(x)).</math>
 |-
 |Using the Chain Rule, we have
@@ 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(g(x)).</math>
 |-
 |Using the Chain Rule, we have

@@ Line 32: / Line 32: @@
 !Solution: &nbsp;
 |-
-|Using the Product Rule, we have
+|Let &nbsp;<math>f(x)=\sin (x)</math>&nbsp; and &nbsp;<math>g(x)=3x.</math>
 |-
+|Then, &nbsp;<math>f'(x)=\cos(x)</math>&nbsp; and &nbsp;<math>g'(x)=3.</math>
 |-
-|Then, using the Power Rule, we have
+|Now, &nbsp;<math>h(x)=f(g(x)).</math>
 |-
+|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'(g(x))\cdot g'(x)}\\
 &&\\
-& = & \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>
 |}
@@ 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>
 |}
@@ Line 83: / Line 61: @@
 !Solution: &nbsp;
 |-
+|Let &nbsp;<math>f(x)=x^8</math>&nbsp; and &nbsp;<math>g(x)=x+1.</math>
 |-
+|Then, &nbsp;<math>f'(x)=8x^7</math>&nbsp; and &nbsp;<math>g'(x)=1.</math>
 |-
-|<u>NOTE:</u> It is not necessary to use the Quotient Rule to calculate the derivative of this function.
+|Now, &nbsp;<math>h(x)=f(g(x)).</math>
 |-
-|You can divide and then use the Power Rule.
+|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'(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>
 |}
@@ 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>
 |}
@@ Line 132: / Line 90: @@
 !Solution: &nbsp;
 |-
-|Using the Quotient Rule, we get
+|Let &nbsp;<math>f(x)=\ln (x)</math>&nbsp; and &nbsp;<math>g(x)=x^2.</math>
 |-
+|
 ::<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'(g(x))\cdot g'(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>
 |}
@@ 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 25: / Line 25: @@
 ==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;"
@@ Line 78: / Line 78: @@
 |}
-'''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;"
@@ Line 127: / Line 127: @@
 |}
-'''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;"

@@ Line 4: / Line 4: @@
 But, what about more complicated functions?
-For example, &nbsp;<math>f(x)=\sin(3x)</math>&nbsp; or &nbsp;<math>g(x)=(x+1)^8?</math>
+For example, &nbsp;<math style="vertical-align: -5px">f(x)=\sin(3x)</math>&nbsp; or &nbsp;<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 &nbsp;<math>f(x)=\sin(3x),</math>&nbsp; it is the composition of the function &nbsp;<math>y=3x</math>&nbsp; with &nbsp;<math>y=\sin(x).</math>
+For &nbsp;<math style="vertical-align: -5px">f(x)=\sin(3x),</math>&nbsp; it is the composition of the function &nbsp;<math style="vertical-align: -4px">y=3x</math>&nbsp; with &nbsp;<math style="vertical-align: -5px">y=\sin(x).</math>
-Similarly, for &nbsp;<math>g(x)=(x+1)^8,</math>&nbsp; it is the composition of &nbsp;<math>y=x+1</math>&nbsp; and &nbsp;<math>y=x^8.</math>
+Similarly, for &nbsp;<math style="vertical-align: -5px">g(x)=(x+1)^8,</math>&nbsp; it is the composition of &nbsp;<math style="vertical-align: -5px">y=x+1</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">y=x^8.</math>
 So, how do we take the derivative of compositions?
@@ Line 18: / Line 18: @@
 '''Chain Rule'''
-Let &nbsp;<math>y=f(u)</math>&nbsp; be a differentiable function of &nbsp;<math>u</math>&nbsp; and let &nbsp;<math>u=g(x)</math>&nbsp; be a differentiable function of &nbsp;<math>x.</math>&nbsp;
+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>y=f(g(x))</math>&nbsp; is a differentiable function of &nbsp;<math>x</math>&nbsp; and
+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
 ::<math>y'=f'(g(x))\cdot g'(x).</math>
 ==Warm-Up==

← Older revision		Revision as of 16:14, 9 October 2017
Line 1:		Line 1:
	==Introduction==		==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>f(x)=\sin(3x)</math>  or  <math>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>f(x)=\sin(3x),</math>  it is the composition of the function  <math>y=3x</math>  with  <math>y=\sin(x).</math>
		+
		+	Similarly, for  <math>g(x)=(x+1)^8,</math>  it is the composition of  <math>y=x+1</math>  and  <math>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>y=f(u)</math>  be a differentiable function of  <math>u</math>  and let  <math>u=g(x)</math>  be a differentiable function of  <math>x.</math>
		+
		+	Then,  <math>y=f(g(x))</math>  is a differentiable function of  <math>x</math>  and
		+
		+	::<math>y'=f'(g(x))\cdot g'(x).</math>
		+
		+
		+
	Taking the derivatives of <em>simple functions</em> (i.e. polynomials) is easy using the power rule.		Taking the derivatives of <em>simple functions</em> (i.e. polynomials) is easy using the power rule.