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Calculus 2 : Calculus II

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

Example Questions

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 #2 : 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 #3 : Derivative At A Point

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:

$\small 0$

$\small -2$

$\small -1$

$\small 1$

$\small 2$

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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Example Question #1 : Derivative At A Point

Find the derivative of the following function at the point $x=\frac{\pi}2$ .

f(x)=sin(x)ln(x)

Possible Answers:

$\frac{\pi}{2}$

0.45

$\frac{2}{\pi}$

Correct answer:

$\frac{2}{\pi}$

Explanation:

Here, we must use the product rule.

Assuming f(x)=sin(x) and g(x)= ln(x) , our expression becomes

$cos(x)ln(x)+\frac{sin(x)}{x}$ .

The question asks us to evaluate this at $x=\frac{\pi}2$ .

$cos\left(\frac{\pi}{2}\right)ln\left(\frac{\pi}{2}\right)+\frac{sin(\frac{\pi}{2})}{\frac{\pi}{2}}=\frac{2}{\pi}$

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Example Question #1 : Derivative At A Point

Given the function $f(x)=2x^{2}-4x+7$ , what is the slope at the point (5,10) ?

Possible Answers:

Correct answer:

Explanation:

As slope is defined as the derivative of a given function at a given point, we will need to take the derivative of $f(x)=2x^{2}-4x+7$ and substitute in the -value of the point (5,10) .

Using the Power Rule $\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$ , $f'(x)=(2)2x^{2-1}-(1)4x^{1-1}+(0)7=4x-4$ . Subbing in x=5 , f'(5)=4(5)-4=20-4=16 .

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Example Question #91 : Derivative Review

Given the function $f(x)=5x^{2}+9x-12$ , what is the slope at the point (-1,3) ?

Possible Answers:

Correct answer:

Explanation:

As slope is defined as the derivative of a given function at a given point, we will need to take the derivative of $f(x)=5x^{2}+9x-12$ and substitute in the -value of the point (-1,3) .

Using the Power Rule $\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$ , $f'(x)=(2)5x^{2-1}+(1)9x^{1-1}-(0)12=10x+9$ , Subbing in x=-1 , we get f'(-1)=10(-1)+9=-10+9=-1 .

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Example Question #92 : Derivative Review

Given the function $f(x)=x^{2}-7x-6$ , what is the slope at the point (2,9) ?

Possible Answers:

Correct answer:

Explanation:

As slope is defined as the derivative of a given function at a given point, we will need to take the derivative of $f(x)=x^{2}-7x-6$ and substitute in the -value of the point (2,9) .

Using the Power Rule $\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$ ,

$f'(x)=(2)x^{2-1}-(1)7x^{1-1}-(0)6=2x-7$ .

Swapping in x=2 , we get f'(2)=2(2)-7=4-7=-3 .

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Example Question #93 : Derivative Review

Given $f(x)=7x^{2}-2x+9$ , find the value of f'(x) at the point (2,5) .

Possible Answers:

Correct answer:

Explanation:

Given the function $f(x)=7x^{2}-2x+9$ , we can use the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$ to find its derivative:

$f'(x)=(2)7x^{2-1}-(1)2x^{1-1}+(0)9=14x-2$ .

Plugging in the -value of the point (2,5) into f'(x) , we get

f'(2)=14(2)-2=28-2=26 .

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #1 : Derivative At A Point

Example Question #1 : Derivative At A Point

Example Question #2 : Derivative At A Point

Example Question #1 : Derivative At A Point

Example Question #3 : Derivative At A Point

Example Question #1 : Derivative At A Point

Example Question #1 : Derivative At A Point

Example Question #91 : Derivative Review

Example Question #92 : Derivative Review

Example Question #93 : Derivative Review

All Calculus 2 Resources

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