Product Rule and Quotient Rule - Revision history

Grad: /* Introduction */

2017-10-05T05:46:23Z

Introduction

Kayla Murray: /* Exercise 2 */

2017-10-04T17:09:55Z

Exercise 2

Kayla Murray: /* Exercise 2 */

2017-10-04T17:08:36Z

Exercise 2

Kayla Murray: /* Exercise 1 */

2017-10-04T17:08:06Z

Exercise 1

Kayla Murray: /* Exercise 4 */

2017-10-04T17:01:41Z

Exercise 4

Kayla Murray: /* Exercise 3 */

2017-10-04T16:54:09Z

Exercise 3

Kayla Murray: /* Exercise 2 */

2017-10-04T16:46:44Z

Exercise 2

Kayla Murray: /* Exercise 1 */

2017-10-04T16:37:01Z

Exercise 1

Kayla Murray: /* Exercise 4 */

2017-09-28T16:59:02Z

Exercise 4

Kayla Murray: /* Exercise 3 */

2017-09-28T16:57:36Z

Exercise 3

@@ Line 10: / Line 10: @@
 Or what about &nbsp;<math style="vertical-align: -5px">g'(x)</math>&nbsp; when &nbsp;<math style="vertical-align: -15px">g(x)=\frac{x}{x+1}?</math>
-Notice &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is a product and &nbsp;<math style="vertical-align: -5px">g(x)</math>&nbsp; is a quotient. So, to answer the question of how to calculate these derivatives, we look to the Product Rule and the Quotient Rule. The Product Rule and the Quotient Rule give us formulas for calculating these derivatives.
+Notice &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is a product, and &nbsp;<math style="vertical-align: -5px">g(x)</math>&nbsp; is a quotient. So, to answer the question of how to calculate these derivatives, we look to the Product Rule and the Quotient Rule. The Product Rule and the Quotient Rule give us formulas for calculating these derivatives.
 '''Product Rule'''

@@ Line 199: / Line 199: @@
 == Exercise 2 ==
-Calculate the derivative of &nbsp;<math style="vertical-align: -3px">g(x)=2x\sin x \sec x.</math>
+Calculate the derivative of &nbsp;<math style="vertical-align: -5px">g(x)=2x\sin x \sec x.</math>
-Notice that the function &nbsp;<math style="vertical-align: -3px">g(x)</math>&nbsp; is the product of three functions.
+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: -3px">g(x)=(2x\sin x)\sec x.</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 Product Rule, we get

@@ Line 199: / Line 199: @@
 == Exercise 2 ==
-Calculate the derivative of &nbsp;<math style="vertical-align: -17px">g(x)=2x\sin x \sec x.</math>
+Calculate the derivative of &nbsp;<math style="vertical-align: -3px">g(x)=2x\sin x \sec x.</math>
 Notice that the function &nbsp;<math style="vertical-align: -3px">g(x)</math>&nbsp; is the product of three functions.

@@ Line 164: / Line 164: @@
 == Exercise 1 ==
-Calculate the derivative of &nbsp;<math style="vertical-align: -16px">f(x)=\frac{1}{x^2}(\csc x-4).</math>
+Calculate the derivative of &nbsp;<math style="vertical-align: -13px">f(x)=\frac{1}{x^2}(\csc x-4).</math>
-First, we need to know the derivative of &nbsp;<math style="vertical-align: -1px">\csc x.</math>&nbsp; Recall
+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>

@@ Line 259: / Line 259: @@
 Calculate the derivative of  &nbsp;<math style="vertical-align: -14px">f(x)=\frac{e^x}{x^2\sin x}.</math>
-Let &nbsp;<math style="vertical-align: -2px">u=x+2.</math>&nbsp; Then, &nbsp;<math style="vertical-align: -1px">du=dx.</math>
+First, using the Quotient Rule, we have
-Now, we need a way of replacing &nbsp;<math style="vertical-align: -2px">x^2+4.</math>
+::<math>\begin{array}{rcl}
-If we solve for &nbsp;<math style="vertical-align: 0px">x</math>&nbsp; in our first equation, we get &nbsp;<math style="vertical-align: -1px">x=u-2.</math>
+Now, we need to use the Product Rule. So, we have
 ::<math>\begin{array}{rcl}
-\displaystyle{\int \frac{x^2+4}{x+2}~dx} & = & \displaystyle{\int \frac{(u-2)^2+4}{u}~du}\\
+\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{\frac{(x+2)^2}{2}-4(x+2)+8\ln|x+2|+C.}\\
+& = & \displaystyle{\frac{x^2\sin x e^x - e^x(x^2\cos x+2x\sin x)}{x^4\sin^2 x}.}
 \end{array}</math>
 So, we have
-::<math>\int \frac{x^2+4}{x+2}~dx=\frac{(x+2)^2}{2}-4(x+2)+8\ln|x+2|+C.</math>
+::<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>

@@ Line 240: / Line 240: @@
 Calculate the derivative of &nbsp;<math style="vertical-align: -16px">h(x)=\frac{x^2\sin x+1}{x^2\cos x+3}.</math>
-Here, the substitution is not obvious.
+Using the Quotient Rule, we have
-Let &nbsp;<math style="vertical-align: -3px">u=2x+3.</math>&nbsp; Then, &nbsp;<math style="vertical-align: -1px">du=2~dx</math>&nbsp; and &nbsp;<math style="vertical-align: -14px">\frac{du}{2}=dx.</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>
-Now, we need a way of getting rid of &nbsp;<math style="vertical-align: -2px">x+5</math>&nbsp; in the numerator.
+Now, we need to use the Product Rule. So, we have
 ::<math>\begin{array}{rcl}
-\displaystyle{\int \frac{x+5}{2x+3}~dx} & = & \displaystyle{\int \frac{(\frac{1}{2}u-\frac{3}{2})+5}{2u}~du}\\
+\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{\frac{1}{4}(2x+3+7\ln|2x+3|)+C.}\\
+& = & \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}.}
 \end{array}</math>
 So, we get
-::<math>\int \frac{x+5}{2x+3}~dx=\frac{1}{4}(2x+3+7\ln|2x+3|)+C.</math>
+::<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>
 == Exercise 4 ==

@@ Line 201: / Line 201: @@
 Calculate the derivative of &nbsp;<math style="vertical-align: -17px">g(x)=2x\sin x \sec x.</math>
-Let &nbsp;<math style="vertical-align: -5px">u=5+\sin(x).</math>&nbsp; Then, &nbsp;<math style="vertical-align: -5px">u=\cos(x)~dx.</math>
+Notice that the function &nbsp;<math style="vertical-align: -3px">g(x)</math>&nbsp; is the product of three functions.
-Plugging these into our integral, we get
+We start by grouping two of the functions together. So, we have &nbsp;<math style="vertical-align: -3px">g(x)=(2x\sin x)\sec x.</math>
 ::<math>\begin{array}{rcl}
-\displaystyle{\int \frac{\cos(x)}{(5+\sin x)^2}~dx} & = & \displaystyle{\int \frac{1}{u^2}~du}\\
+\displaystyle{g'(x)} & = & \displaystyle{(2x\sin x)(\sec x)'+(2x\sin x)'\sec x}\\
 &&\\
-& = & \displaystyle{-\frac{1}{u}+C}\\
+& = & \displaystyle{(2x\sin x)(\tan^2 x)+(2x\sin x)'\sec x.}
 &&\\
-& = & \displaystyle{-\frac{1}{5+\sin(x)}+C.}
+& = & \displaystyle{2x\sin x\tan^2 x+(2x\cos x+2\sin x)\sec x.}
 \end{array}</math>
 So, we have
-::<math>\int \frac{\cos(x)}{(5+\sin x)^2}~dx=-\frac{1}{5+\sin(x)}+C.</math>
+::<math>g'(x)=2x\sin x\tan^2 x+(2x\cos x+2\sin x)\sec x.</math>
 == Exercise 3 ==

@@ Line 166: / Line 166: @@
 Calculate the derivative of &nbsp;<math style="vertical-align: -16px">f(x)=\frac{1}{x^2}(\csc x-4).</math>
-First, we factor out &nbsp;<math style="vertical-align: -1px">4</math>&nbsp; out of the denominator.
+First, we need to know the derivative of &nbsp;<math style="vertical-align: -1px">\csc x.</math>&nbsp; Recall
-So, we have
+::<math>\csc x =\frac{1}{\sin x}.</math>
 ::<math>\begin{array}{rcl}
-\displaystyle{\int \frac{2}{y^2+4}~dy} & = & \displaystyle{\frac{1}{4}\int \frac{2}{\frac{y^2}{4}+1}~dy}\\
+\displaystyle{\frac{d}{dx}(\csc x)} & = & \displaystyle{\frac{d}{dx}\bigg(\frac{1}{\sin x}\bigg)}\\
 &&\\
-& = & \displaystyle{\frac{1}{2}\int \frac{1}{(\frac{y}{2})^2+1}~dy.}\\
+& = & \displaystyle{\frac{-\cos x}{\sin^2 x}} \\
 \end{array}</math>
-Now, we use &nbsp;<math style="vertical-align: -1px">u</math>-substitution. Let &nbsp;<math>u=\frac{y}{2}.</math>
+Using the Product Rule and Power Rule, we have
 ::<math>\begin{array}{rcl}
-\displaystyle{\int \frac{2}{y^2+4}~dy} & = & \displaystyle{\frac{1}{2}\int \frac{2}{u^2+1}~du}\\
+\displaystyle{f'(x)} & = & \displaystyle{\frac{1}{x^2}(\csc x-4)'+\bigg(\frac{1}{x^2}\bigg)'(\csc x-4)}\\
 &&\\
-& = & \displaystyle{\arctan(u)+C}\\
+& = & \displaystyle{\frac{1}{x^2}(-\csc x \cot x+0)+(-2x^{-3})(\csc x-4)}\\
 &&\\
-& = & \displaystyle{\arctan\bigg(\frac{y}{2}\bigg)+C.}\\
+& = & \displaystyle{\frac{-\csc x \cot x}{x^2}+\frac{-2(\csc x-4)}{x^3}.}
 \end{array}</math>
 So, we have
-::<math>\int \frac{2}{y^2+4}~dy=\arctan\bigg(\frac{y}{2}\bigg)+C.</math>
+::<math>f'(x)=\frac{-\csc x \cot x}{x^2}+\frac{-2(\csc x-4)}{x^3}.</math>
 == Exercise 2 ==