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Calculus 2 : Derivative Review

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Example Questions

Example Question #4 : Derivative At A Point

What is the equation of the line tangent to the graph of the function

$f(x) = x^{3} + x - 17$

at the point (3, 13) ?

Possible Answers:

y = 28 x- 71

y = 26 x- 65

y = 10x -17

y = 11x -20

y = 27 x- 68

Correct answer:

y = 28 x- 71

Explanation:

The slope of the line tangent to the graph of $f(x) = x^{3} + x - 17$ at the point (3, 13) is f'(3) , which can be evaluated as follows:

$f(x) = x^{3} + x - 17$

$f'(x) = \frac{\mathrm{d} }{\mathrm{d} x}\left ( x^{3} + x - 17 \right )$

$f'(x) = 3x^{2} + 1$

$f'(3) = 3\cdot 3^{2} + 1 = 28$

The line with slope 28 through (3, 13) has equation:

$y - y_{1} = m (x-x_{1})$

y -13= 28 (x-3)

$y -13= 28 x- 28 \cdot 3$

y -13= 28 x- 84

y = 28 x- 71

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Example Question #1195 : Calculus Ii

Let the initial approximation of the ninth root of 100 be

$x_{1}= 2$ .

Use one iteration of Newton's method to find approximation $x_{2}$ . Give your answer to the nearest thousandth.

Possible Answers:

1.821

1.971

1.921

2.021

1.871

Correct answer:

1.821

Explanation:

This is equivalent to finding a solution of the equation

x^9 = 100 ,

or the zero of the polynomial

P (x) = x^9 - 100 .

Using Newton's method, we can find $x_{2}$ from the formula

$x_{2} = x_{1} - \frac{P(x_{1})}{P'(x_{1})}$ .

P '(x) = 9 x^8 , so

$P (x_{1}) = P(2)= 2^9 - 100 = 512 - 100 = 412$

$P '(x_{1}) = P' (2)= 9 \cdot 2^8 = 9 \cdot 256 = 2,304$

$x_{2} = 2 - \frac{P(2)}{P'(2)} = 2 - \frac{412}{2,304} \approx 2 - 0.1788 \approx 1.821$

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Example Question #4 : Chain Rule And Implicit Differentiation

$g(x) = e ^{x^{2}+x}$

Evaluate g '(3 ) .

Possible Answers:

$7 e^{9}$

$e^{7}$

$12 e^{12}$

$7 e^{12}$

$12 e^{11}$

Correct answer:

$7 e^{12}$

Explanation:

To find g'(x) , substitute $u = x ^{2} + x$ and use the chain rule:

$g(x) = e ^{x^{2}+x}$

$g'(x) = \frac{\mathrm{d} }{\mathrm{d} x} e ^{x^{2}+x}$

$= \frac{\mathrm{d} }{\mathrm{d} x} e ^{u}$

$= \frac{\mathrm{d}u }{\mathrm{d} x} \cdot \frac{\mathrm{d} }{\mathrm{d} u}e ^{u}$

$= \frac{\mathrm{d}u }{\mathrm{d} x} \cdot e ^{u}$

$= \frac{\mathrm{d} }{\mathrm{d} x}\left ( x ^{2} + x \right ) \cdot \frac{\mathrm{d} }{\mathrm{d} u}e ^{u}$

$= \left ( 2x + 1 \right ) \cdot e ^{u}$

$= \left ( 2x + 1 \right ) \cdot e ^{ x ^{2} + x}$

$g'(x)= \left ( 2x + 1 \right ) e ^{ x ^{2} + x}$

Plug in 3:

$g'(3)= \left ( 2 \cdot 3 + 1 \right ) \cdot e ^{ 3 ^{2} + 3}$

$= \left ( 6 + 1 \right ) \cdot e ^{ 9+ 3} =7e ^{ 12}$

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Example Question #5 : Chain Rule And Implicit Differentiation

$g (x) = \frac{1}{4x - 7}$

Evaluate g ' (1 ) .

Possible Answers:

$- \frac{4}{9}$

$-\frac{4}{3}$

$\frac{4}{9}$

Undefined

$\frac{4}{3}$

Correct answer:

$- \frac{4}{9}$

Explanation:

To find g'(x) , substitute u =4x - 7 and use the chain rule:

$g (x) = \frac{1}{4x - 7}$

$g '(x) = \frac{\mathrm{d} }{\mathrm{d} x}\cdot \frac{1}{4x - 7}$

$= \frac{\mathrm{d} }{\mathrm{d} x}\cdot \frac{1}{u}$

$= \frac{\mathrm{d} u}{\mathrm{d} x}\cdot \frac{\mathrm{d} }{\mathrm{d} u} \frac{1}{u}$

$= \frac{\mathrm{d} }{\mathrm{d} x} (4x-7) \cdot \frac{\mathrm{d} }{\mathrm{d} u} u ^{-1}$

$=4 \cdot (-1 ) u ^{-1-1}$

$=-4 u ^{-2} = - \frac{4}{u^{2}}= - \frac{4}{(4x-7)^{2}}$

$g'(x)= - \frac{4}{(4x-7)^{2}}$

and

$g'(1)= - \frac{4}{(4 \cdot 1 -7)^{2}}$

$= - \frac{4}{(4 -7)^{2}}$

$= - \frac{4}{(-3)^{2}} = - \frac{4}{9}$

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Example Question #6 : Chain Rule And Implicit Differentiation

$g (x) = \tan \left ( x +\frac{ \pi }{4}\right )$

Evaluate $g' \left ( \pi )$ .

Possible Answers:

$\sqrt{2}$

$-\sqrt{2}$

Undefined

Correct answer:

Explanation:

To find g'(x) , substitute $u = x +\frac{ \pi }{4}$ and use the chain rule:

$g (x) = \tan \left ( x +\frac{ \pi }{4}\right )$

$g' (x) = \frac{\mathrm{d} }{\mathrm{d} x}\tan \left ( x +\frac{ \pi }{4}\right )$

$g' (x) = \frac{\mathrm{d} }{\mathrm{d} x}\tan u$

$=\frac{\mathrm{d} u}{\mathrm{d} x} \cdot \frac{\mathrm{d} }{\mathrm{d} u} \tan u$

$=\frac{\mathrm{d} }{\mathrm{d} x} \left ( x +\frac{ \pi }{4} \right )\cdot \frac{\mathrm{d} }{\mathrm{d} u} \tan u$

$=\frac{\mathrm{d} }{\mathrm{d} x} \left ( x +\frac{ \pi }{4} \right ) \cdot \sec ^{2} u$

$=1 \cdot \sec ^{2}\left ( x +\frac{ \pi }{4}\right ) = \sec ^{2}\left ( x +\frac{ \pi }{4}\right )$

$g'(x) = \sec ^{2}\left ( x +\frac{ \pi }{4}\right )$

$g'(\pi ) = \sec ^{2}\left ( \pi +\frac{ \pi }{4}\right )$

$g'(\pi ) = \sec ^{2}\frac{ 5 \pi }{4}$

$= \left ( - \sqrt{2 }\right )^{2} = 2$

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Example Question #7 : Chain Rule And Implicit Differentiation

$g (x) = 3 ^{x^{2}}$

Evaluate g ' (-1 ) .

Possible Answers:

$-\frac{3 \ln 3}{2}$

$-6 \ln 3$

Correct answer:

$-6 \ln 3$

Explanation:

To find g'(x) , substitute $u = x ^{2}$ and use the chain rule:

$g (x) = 3 ^{x^{2}}$

$g ' (x) = \frac{\mathrm{d} }{\mathrm{d} x} 3 ^{x^{2}}$

$= \frac{\mathrm{d} }{\mathrm{d} x} 3 ^{u}$

$= \frac{\mathrm{d}u }{\mathrm{d} x} \cdot \frac{\mathrm{d} }{\mathrm{d} u} 3 ^{u}$

$= \frac{\mathrm{d}}{\mathrm{d} x} x^{2} \cdot \frac{\mathrm{d} }{\mathrm{d} u} 3 ^{u}$

$= 2x \cdot \ln 3 \cdot 3 ^{u}$

$= 2x \ln 3 \cdot 3 ^{x^2}$

So $g'(x)= 2x \ln 3 \cdot 3 ^{x^2}$

and

$g'(-1)= 2(-1) \ln 3 \cdot 3 ^{(-1)^2}$

$= -2 \ln 3 \cdot 3 ^{1} = -2 \ln 3 \cdot 3 = -6 \ln 3$

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Example Question #8 : Chain Rule And Implicit Differentiation

$g \left (x \right )= \cos \left ( \pi x^{2}\right )$

Evaluate $g ' \left ( \frac{1}{2} \right )$ .

Possible Answers:

$\frac{ \pi \sqrt{2}}{2}$

$- \frac{ \sqrt{2}}{2 \pi}$

$- \pi \sqrt{2}$

$- \frac{ \pi \sqrt{2}}{2}$

$\pi \sqrt{2}$

Correct answer:

$- \frac{ \pi \sqrt{2}}{2}$

Explanation:

To find g'(x) , substitute $u = \pi x^{2}$ and use the chain rule:

$g \left (x \right )= \cos \left ( \pi x^{2}\right )$

$g'(x)= \frac{\mathrm{d} }{\mathrm{d} x} \cos \left ( \pi x^{2}\right )$

$= \frac{\mathrm{d} }{\mathrm{d} x} \cos u$

$=\frac{\mathrm{d} u}{\mathrm{d} x} \cdot \frac{\mathrm{d} }{\mathrm{d} u} \cos u$

$=\frac{\mathrm{d} }{\mathrm{d} x} \pi x^{2} \cdot \frac{\mathrm{d} }{\mathrm{d} u} \cos u$

$= \pi \cdot 2 x\cdot \left ( - \sin u \right )$

$= \pi \cdot 2 x\cdot \left [- \sin \left ( \pi x^{2} ) \right ]$

$=-2 \pi x \sin \left ( \pi x^{2} )$

So $g'(x)=-2 \pi x \sin \left ( \pi x^{2} )$

and

$g ' \left ( \frac{1}{2} \right )=-2 \pi\cdot \frac{1}{2} \sin \left [\pi \cdot \left ( \frac{1}{2} \right )^{2} \right ]$

$=- \pi \sin \left ( \frac{\pi}{4} \right )$

$=- \pi \cdot \frac{\sqrt{2}}{2} = - \frac{ \pi \sqrt{2}}{2}$

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Example Question #71 : Derivative Review

What is the equation of the line tangent to the graph of the function

f(x)= x^3-2

at (2,6) ?

Possible Answers:

y = 10 x-14

y = 24 x-42

y = 12 x-18

y = 12 x+18

y = 10 x+14

Correct answer:

y = 12 x-18

Explanation:

The slope of the line tangent to the graph of f(x)= x^3-2 at (2,6) is

f'(2) , which can be evaluated as follows:

f(x)= x^3-2

$f'(x)= 3x^{3-1}-0$

$f'(x)= 3x^{2}$

Then $f'(2)= 3 \cdot 2^{2} = 3 \cdot4 = 12$ , which is the slope of the line.

The equation of the line with slope 12 through (2,6) is:

$y - y_{1} = m (x-x_{1})$

y - 6 = 12 (x-2)

y - 6 = 12 x-24

y - 6 + 6= 12 x-24 + 6

y = 12 x-18

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Example Question #71 : Derivatives

Let the initial approximation of the seventh root of 1,000 be

$x_{1}= 3$ .

Use one iteration of Newton's method to find approximation $x_{2}$ . Give your answer to the nearest thousandth.

Possible Answers:

2.727

2.787

2.767

2.707

2.687

Correct answer:

2.767

Explanation:

This is equivalent to finding a solution of the equation

x^7 = 1,000 ,

or the zero of the polynomial

P (x) = x^7 - 1,000 .

Using Newton's method, we can find $x_{2}$ from the formula

$x_{2} = x_{1} - \frac{P(x_{1})}{P'(x_{1})}$ .

P'(x) = 7x^6 , so

$P (x_{1}) =P(3)= 3^7 - 1,000 = 2,187 - 1,000 = 1,187$

$P'(x_{1}) = P'(3)= 7 \cdot 3^6 = 7 \cdot 729 = 5,103$

$x_{2} = 3 - \frac{P(3)}{P'(3)} = 3 - \frac{1,187}{5,103} \approx 3 - 0.2326 \approx 2.767$

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Example Question #1203 : Calculus Ii

Let the initial approximation of a solution of the equation

x^3 = x + 19

be $x_{1} = 2.5$ .

Use one iteration of Newton's method to find an approximation of $x_{2}$ . Give your answer to the nearest thousandth.

Possible Answers:

2.731

2.681

2.831

2.881

2.781

Correct answer:

2.831

Explanation:

Rewrite the equation to be solved for as

x^3 - x - 19 = 0 .

Let f(x) = x^3 - x - 19 .

Then,

$f'(x) = \frac{\mathrm{d} }{\mathrm{d} x} \left ( x^3 - x - 19 \right )$ .

f'(x) =3x^2 - 1

The problem amounts to finding a zero of f(x ) . By Newton's method, the second approximation can be derived from the first using the equation

$x_{2} = x_{1} - \frac{f(x_{1})}{f'(x_{1})}$ .

$x_{1} = 2.5$ , so

f(2.5) = 2.5^3 - 2.5 - 19 = 15.625 - 2.5 - 19 = -5.875

and

$f'(2.5) =3 \cdot 2.5^2 - 1 = 3 \cdot 6.25 - 1 = 18.75 - 1 = 17.75$

$x_{2} = 2.5 - \frac{f(2.5)}{f'(2.5)}= 2.5 - \frac{ -5.875}{17.75}\approx 2.5 - (-0.3310) \approx 2.831$

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Calculus 2 : Derivative Review

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #4 : Derivative At A Point

Example Question #1195 : Calculus Ii

Example Question #4 : Chain Rule And Implicit Differentiation

Example Question #5 : Chain Rule And Implicit Differentiation

Example Question #6 : Chain Rule And Implicit Differentiation

Example Question #7 : Chain Rule And Implicit Differentiation

Example Question #8 : Chain Rule And Implicit Differentiation

Example Question #71 : Derivative Review

Example Question #71 : Derivatives

Example Question #1203 : Calculus Ii

All Calculus 2 Resources

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