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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

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9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #4 : Chain Rule And Implicit Differentiation

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

Evaluate g ' (-1 ) .

Possible Answers:

$-6 \ln 3$

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

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 #1201 : Calculus Ii

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

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

Possible Answers:

$\pi \sqrt{2}$

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

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

$- \pi \sqrt{2}$

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

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 #1202 : Calculus Ii

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 #73 : 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

2.707

2.767

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 #71 : Derivatives

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:

2.881

2.681

2.781

2.831

2.731

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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Example Question #1203 : Calculus Ii

Let the initial approximation of a solution of the equation

$x + 2 \ln x = 7$

be $x_{1} = 4$ .

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

Possible Answers:

4.052

4.002

3.952

4.152

4.102

Correct answer:

4.152

Explanation:

Rewrite the equation to be solved for as

$x + 2 \ln x - 7 = 0$ .

We are therefore trying to find a zero of the function

$f(x) = x + 2 \ln x - 7$ .

$f'(x) = \frac{\mathrm{d} }{\mathrm{d} x}\left ( x + 2 \ln x - 7 \right ) = 1 + 2\cdot \frac{1}{x} = 1 + \frac{2}{x}$

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

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

$f(x_{1}) = f (4) = 4 + 2 \ln4 - 7$

$= - 3 + \ln 16$

$\approx -3 + 2.7726 \approx -0.2274$

$f'(x_{1}) = f (4) = 1 + \frac{2}{4} = 1.5$

$x_{2} \approx 4 - \frac{-0.2274}{1.5} \approx 4 + 0.1516 \approx 4.152$

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Example Question #81 : Derivatives

Let the initial approximation of a solution of the equation

$x = \cos 2x$

be $x_{1}= 0.6$ .

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

Possible Answers:

0.517

0.603

0.643

0.557

0.683

Correct answer:

0.517

Explanation:

Rewrite the equation to be solved for as

$x - \cos 2x = 0$ .

We are therefore trying to find a zero of the function

$f(x) = x - \cos 2x$ .

$f'(x) =\frac{\mathrm{d} }{\mathrm{d} x}\left ( x - \cos 2x \right ) = 1 + 2 \sin 2x$ .

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

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

$f(x_{1}) = f (0.6) =0.6 - \cos \left (2 \cdot 0.6 \right )$

$=0.6 - \cos 1.2$

$\approx 0.6 - 0.3624 \approx 0.2376$

$f'(x_{1}) = f'(0.6)= 1 + 2 \sin \left (2 \cdot 0.6 \right )$

$f'(x_{1}) = f'(0.6)= 1 + 2 \sin1.2$

$= 1 + 2 \cdot 0.9320$

$\approx 1 + 2 \cdot 0.9320 \approx 2.8641$

$x_{2} \approx 0.6 - \frac{0.2376}{2.8641} \approx 0.6 - 0.0830 \approx 0.517$

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Example Question #83 : Derivative Review

Given the function y=log_e(x)+5x , find the slope of the point (10,51) .

Possible Answers:

$\frac{51}{10}$

The slope cannot be determined.

Correct answer:

$\frac{51}{10}$

Explanation:

To find the slope at a point of a function, take the derivative of the function.

y=log_e(x)+5x

The derivative of log_a(x) is $\frac{1}{x\:ln(a)}$ .

Therefore the derivative becomes,

$\frac{1}{x\:ln(e)}+5=\frac{1}{x}+5$ since ln(e)=1 .

Now we substitute the given point to find the slope at that point.

$\frac{dy}{dx}=\frac{1}{x} +5 = \frac{1}{10}+5 =\frac{51}{10}$

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Example Question #1 : Derivative At A Point

Find the value of the following derivative at the point $\left(2,\frac{1}{2}\right)$ :

$f(x)=\frac{1}{\sqrt{x+2}}$

Possible Answers:

$\frac{1}{16}$

$\frac{1}{4}$

$-\frac{1}{16}$

$\frac{1}{\sqrt{\frac{5}{2}}}$

${}-\frac{1}{4}$

Correct answer:

$-\frac{1}{16}$

Explanation:

To solve this problem, first we need to take the derivative of the function. It will be easier to rewrite the equation as $f(x)=(x+2)^{-\frac{1}{2}}$ from here we can take the derivative and simplify to get

$\\f'(x)=-\frac{1}{2}(x+2)^{-\frac{1}{2}-1}\cdot \frac{\mathrm{d} }{\mathrm{d} x}(x+2)\\ f'(x)=-\frac{1}{2}(x+2)^{-\frac{3}{2}}\cdot 1\\f'(x)=-\frac{1}{2}(x+2)^{-\frac{3}{2}}$

From here we need to evaluate at the given point $\left(2,\frac{1}{2}\right)$ . In this case, only the x value is important, so we evaluate our derivative at x=2 to get $f'(2)=-\frac{1}{2}(2+2)^{-\frac{3}{2}}=-\frac{1}{2}\cdot \frac{1}{8}=-\frac{1}{16}$ .

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Example Question #2 : Derivative At A Point

Evaluate the value of the derivative of the given function at the point (-1,4) :

f(x)=3x^4+2x^2-1

Possible Answers:

Correct answer:

Explanation:

To solve this problem, first we need to take the derivative of the function.

$\\f'(x)=4\cdot 3x^{4-1}+2\cdot 2x^{2-1}\\f'(x)=12x^3+4x$

From here we need to evaluate at the given point (-1,4) . In this case, only the x value is important, so we evaluate our derivative at x=1 to get

$\\f'(-1)=12(-1)^3+4(-1)\\ f'(-1)=12(-1)+4(-1)\\ f'(-1)=-16$ .

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #4 : Chain Rule And Implicit Differentiation

Example Question #1201 : Calculus Ii

Example Question #1202 : Calculus Ii

Example Question #73 : Derivatives

Example Question #71 : Derivatives

Example Question #1203 : Calculus Ii

Example Question #81 : Derivatives

Example Question #83 : Derivative Review

Example Question #1 : Derivative At A Point

Example Question #2 : Derivative At A Point

All Calculus 2 Resources

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