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Calculus 2 : Derivatives

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Example Questions

Example Question #81 : Derivative Review

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:

4.152

4.052

3.952

4.002

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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Example Question #1 : Derivative At A Point

Find the value of the derivate of the given function at the point (2,1) :

$f(x)=\frac{x^2+2}{x+4}$

Possible Answers:

$\frac{1}2{}$

$\frac{3}{5}$

Correct answer:

$\frac{1}2{}$

Explanation:

To solve this problem, first we need to take the derivative of the function. To do this we need to use the quotient rule and simplified as follows:

$\\f'(x)=\frac{(x+4)\cdot \frac{\mathrm{d} }{\mathrm{d} x}(x^2+2)-(x^2+2)\cdot \frac{\mathrm{d} }{\mathrm{d} x}(x+4)}{(x+4)^2}\\ f'(x)=\frac{(x+4)\cdot 2x-(x^2+2)\cdot 1}{(x+4)^2}\\f'(x)=\frac{2x^2+8x-x^2-2}{(x+4)^2}\\f'(x)=\frac{x^2+8x-2}{(x+4)^2}$

From here we need to evaluate at the given point (2,1) . In this case, only the x value is important, so we evaluate our derivative at x=2 to get

$\\f'(2)=\frac{(2)^2+8(2)-2}{(2+4)^2}\\ f'(2)=\frac{4+16-2}{36}\\f'(2)=\frac{18}{36}=\frac{1}{2}$ .

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Example Question #1 : Derivative At A Point

Find the slope at (1,0) for the function y=-ln(x) .

Possible Answers:

$-\infty$

$\infty$

Correct answer:

Explanation:

The derivative of y=-ln(x) is:

$y'=-\frac{1}{x}$

Substitute the point at x=1 .

$y'=-\frac{1}{x}=-\frac{1}{1}=-1$

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Example Question #1 : Derivative At A Point

f(x)=ln(x)x^2

What is the slope of f(x) at x=1 ?

Possible Answers:

Correct answer:

Explanation:

In order to find the slope of a function at a certain point, plug in that point into the first derivative of the function. Our first step here is to take the first derivative.

Since we see that f(x) is composed of two different functions, we must use the product rule. Remember that the product rule goes as follows:

f(x)=g(x)h(x)

f'(x)=g'(x)h(x) + g(x)h'(x)

Following that procedure, we set ln(x) equal to g(x) and x^2 equal to h(x) .

$f'(x)= \frac{1}{x}(x^2)+ ln(x)(2x)$ ,

which can be simplified to

f'(x)=x+2xln(x) = x(1+2ln(x)) .

Now plug in 1 to find the slope at x=1.

f'(1)= 1(1+2(0))=1

Remember that ln(1)=0 .

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Example Question #1 : Derivative At A Point

Consider the function: 2y^3=3x^2 . What is the derivative at (3,2) ?

Possible Answers:

$\frac{3}{2}$

$\frac{1}{2}$

$\frac{1}{4}$

$\frac{3}{4}$

$\frac{2}{3}$

Correct answer:

$\frac{3}{4}$

Explanation:

To solve for the derivative of 2y^3=3x^2 , use implicit differentiation which means to take the derivative of each term with respect to the variable in that term.

$6y^2 \cdot \frac{dy}{dx}=6x$

$\frac{dy}{dx}=\frac{6x}{6y^2}$

Substitute the point into the derivative.

$\frac{dy}{dx}=\frac{6x}{6y^2}=\frac{6(3)}{6(2)^2}=\frac{3}{4}$

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Example Question #82 : Derivatives

Use implicit differentiation to find the slope of the tangent line to $\small y^3-lny=x^2$ at the point $\small (1,1)$ .

Possible Answers:

$\small 1$

$\small 2$

$\small 0$

$\small -1$

$\small -2$

Correct answer:

$\small 1$

Explanation:

We must take the derivative $\small \left(\small \frac{\mathrm{d} y}{\mathrm{d} x}\right)$ because that will give us the slope. On the left side we'll get

$\small 3y^2\frac{\mathrm{d}y }{\mathrm{d} x}-(1/y)\frac{\mathrm{d} y}{\mathrm{d} x}$ , and on the right side we'll get $\small 2x$ .

We include the $\small \frac{\mathrm{d} y}{\mathrm{d} x}$ on the left side because $\small y$ is a function of $\small x$ , so its derivative is unknown (hence we are trying to solve for it!).

Now we can factor out a $\small \frac{\mathrm{d} y}{\mathrm{d} x}$ on the left side to get

$\small \small \frac{\mathrm{d} y}{\mathrm{d} x}(3y^2-1/y)=2x$ and divide by $\small 3y^2-1/y$ in order to solve for $\small \frac{\mathrm{d} y}{\mathrm{d} x}$ .

Doing this gives you

$\small \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2x}{3y^2-1/y}$ .

We want to find the slope at $\small (1,1)$ , so we can sub in $\small 1$ for $\small x$ and $\small y$ .

$\small \small \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2(1)}{3(1)^2-1/(1)}=2/2=1$ .

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Calculus 2 : Derivatives

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #81 : Derivative Review

Example Question #81 : Derivatives

Example Question #83 : Derivative Review

Example Question #1 : Derivative At A Point

Example Question #2 : Derivative At A Point

Example Question #1 : Derivative At A Point

Example Question #1 : Derivative At A Point

Example Question #1 : Derivative At A Point

Example Question #1 : Derivative At A Point

Example Question #82 : Derivatives

All Calculus 2 Resources

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