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Calculus 3 : Calculus 3

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #258 : Acceleration

The position at a certain point is given by:

$f(x)=\frac{3}{5}x^{5} - 6x + 5$

What is the acceleration at x=2 ?

Possible Answers:

198

Correct answer:

Explanation:

In order to find the acceleration of a given point, you must first find the derivative of the position function which gives you the velocity function:

$f'(x)=3x^{4}-6$

Then, you differentiate the velocity function to get the acceleration function:

f''(x)=12x^3

Then, find f''(x) when x=2 .

The answer is:

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Example Question #291 : How To Find Acceleration

The position of a certain point is given by the following function:

$p(t)= 2t^{\frac{1}{2}} + 3e^t +4$

What is the acceleration at t=4 ?

Possible Answers:

163.7

170.2

342

165.8

160

Correct answer:

163.7

Explanation:

In order to find the acceleration of a certain point, you must first find the derivative of the position function which gives us the velocity function and then the derivative of the velocity function which gives us the acceleration function: p''(t)= v'(t)= a(t)

In this case, the position function is: $p(t)= 2t^{\frac{1}{2}} + 3e^t +4$

The velocity function is found by taking the derivative of the position function: $v(t)= t^{-\frac{1}{2}} + 3e^t$

The acceleration function is found by taking the derivative of the velocity function: $a(t)= {-\frac{1}{2}}t^{(-\frac{3}{2})} + 3e^t$

Finally, to find the accelaration, substitute t=4 into the acceleration function: $a(4)= {-\frac{1}{2}}(4)^{(-\frac{3}{2})} + 3e^{(4)}$

Therefore, the answer is: 163.7

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Example Question #211 : Calculus 3

$f (x) = 7x^{2} + 6x - 9$

Give f'(x) .

Possible Answers:

f'(x) = 14x

$f'(x) = 14x^{2}+6x$

f'(x) = 14x+12

f'(x) = 7x+6

f'(x) = 14x+6

Correct answer:

f'(x) = 14x+6

Explanation:

$\frac{\mathrm{d} }{\mathrm{d} x} \left (ax^{n} \right )= n\cdot ax^{n-1}$ , and the derivative of a constant is 0, so

$f (x) = 7x^{2} + 6x - 9$

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

$f ' (x) = 14\cdot x^{1} + 6$

f ' (x) = 14x + 6

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Example Question #1 : Computation Of Derivatives

$f(x)= 8x^{3}-7x^{2}+11$

Give f''(x) .

Possible Answers:

f ''(x) = 48x -14

f ''(x) = 48x

f ''(x) = 24x

f ''(x) = 24x - 11

f ''(x) = 24x -14

Correct answer:

f ''(x) = 48x -14

Explanation:

First, find the derivative f'(x) of $f(x)= 8x^{3}-7x^{2}+11$ .

$\frac{\mathrm{d} }{\mathrm{d} x} \left (ax^{n} \right )= n\cdot ax^{n-1}$ , and the derivative of a constant is 0, so

$f(x)= 8x^{3}-7x^{2}+11$

$f'(x)= 3 \cdot 8x^{3-1}-2\cdot 7x^{2-1}+0$

$f'(x)= 24x^{2}-14x^{1}$

$f'(x)= 24x^{2}-14x$

Now, differentiate f'(x) to get f''(x) .

$f''(x)= 2 \cdot 24x^{2-1}-1\cdot 14x^{1-1}$

$f''(x)= 48x^{1}-14x^{0}$

f''(x)= 48x-14

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Example Question #213 : Calculus 3

Differentiate $x^{999}$ .

Possible Answers:

$\frac{\mathrm{d} }{\mathrm{d} x} \left ( x^{999}\right ) = 999x^{998}$

$\frac{\mathrm{d} }{\mathrm{d} x} \left ( x^{999}\right ) = 998x^{998}$

$\frac{\mathrm{d} }{\mathrm{d} x} \left ( x^{999}\right ) = 998x^{1,000}$

$\frac{\mathrm{d} }{\mathrm{d} x} \left ( x^{999}\right ) = 998x^{999}$

$\frac{\mathrm{d} }{\mathrm{d} x} \left ( x^{999}\right ) = 1,000x^{1,000}$

Correct answer:

$\frac{\mathrm{d} }{\mathrm{d} x} \left ( x^{999}\right ) = 999x^{998}$

Explanation:

$\frac{\mathrm{d} }{\mathrm{d} x} \left (x^{n} \right )= n\cdot x^{n-1}$ , so

$\frac{\mathrm{d} }{\mathrm{d} x} \left ( x^{999}\right ) = 999x^{999-1} = 999x^{998}$

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Example Question #211 : Calculus 3

$f(x) =0.9x^{4} + 0.7x^{3}- 0.5 x$

Give f''(x) .

Possible Answers:

$f''(x) =10.8x^{2} + 4.2x$

$f''(x) =10.8x^{2} + 4.2x - 0.5$

$f''(x) =10.8x^{2} + 8.4x$

$f''(x) =10.8x^{2} + 8.4x - 0.5$

$f''(x) =10.8x^{2} + 2.1x$

Correct answer:

$f''(x) =10.8x^{2} + 4.2x$

Explanation:

First, find the derivative f'(x) of $f(x) =0.9x^{4} + 0.7x^{3}- 0.5 x$ .

Recall that $\frac{\mathrm{d} }{\mathrm{d} x} \left (ax^{n} \right )= n\cdot ax^{n-1}$ , and the derivative of a constant is 0.

$f(x) =0.9x^{4} + 0.7x^{3}- 0.5 x$

$f'(x) =4\cdot 0.9x^{4-1} + 3\cdot 0.7x^{3-1}- 1 \cdot 0.5 x ^{1-1}$

$f'(x) =3.6x^{3} + 2.1x^{2}- 0.5 x ^{0}$

$f'(x) =3.6x^{3} + 2.1x^{2}- 0.5$

Now, differentiate f'(x) to get f''(x) .

$f'(x) =3.6x^{3} + 2.1x^{2}- 0.5$

$f''(x) =3 \cdot 3.6x^{3-1} + 2 \cdot 2.1x^{2-1}- 0$

$f''(x) =10.8x^{2} + 4.2x^{1}- 0$

$f''(x) =10.8x^{2} + 4.2x$

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Example Question #1 : First And Second Derivatives Of Functions

Find the second derivative of the following function.

f(x)=sec(x)

Possible Answers:

f''(x)=sec(x)tan(x)

f''(x)=cot(x)csc(x)

f''(x)=-csc(x)

f''(x)=sec^3(x)tan(x)

$f''(x)=sec(x)\cdot (tan^2(x)+sec^2(x))$

Correct answer:

$f''(x)=sec(x)\cdot (tan^2(x)+sec^2(x))$

Explanation:

To find the second derivative, first we need to find the first derivative. So for the given function, we get the first derivative to be

f'(x)=tan(x)sec(x) .

Now we have to take the derivative of the derivative. To do this we need to use the product rule as shown below

Thus, we get

$\\f''(x)=tan(x)\cdot \frac{\mathrm{d} }{\mathrm{d} x}sec(x)+sec(x)\cdot \frac{\mathrm{d} }{\mathrm{d} x}tan(x)\\f''(x)=tan(x)\cdot sec(x)\cdot tan(x)+sec(x)\cdot sec^2(x)\\f''(x)=sec(x)\cdot (tan^2(x)+sec^2(x))$ .

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Example Question #214 : Calculus 3

Find the derivative of $\small lnx^{lnx}$ .

Possible Answers:

$\small lnxlnx/x$

$\small \small lnx^{lnx}(ln(lnx)/x+1/x)$

$\small \small lnx^{lnx}(ln(lnx)+1/x)$

$\small \small lnx^{lnx}ln(lnx)/x$

$\small (lnxlnx^{lnx-1})/x$

Correct answer:

$\small \small lnx^{lnx}(ln(lnx)/x+1/x)$

Explanation:

To solve this derivative, we need to use logarithmic differentiation. This allows us to use the logarithm rule $\small lna^{b}=blna$ to solve an easier derivative.

Let $\small y=lnx^{lnx}$ .

Now we'll take the natural log of both sides to get

$\small lny=ln(lnx^{lnx})=lnxln(lnx)$ .

Now we can use implicit differentiation to solve for $\small y'$ .

The derivative of $\small lny$ is $\small y'/y$ , and the derivative of $\small \small \small lnx(ln(lnx))$ can be found using the product rule, which states

$\small \frac{\mathrm{d} }{\mathrm{d} x}(u\cdot v)=u'v+uv'$ where $\small u$ and $\small v$ are functions of $\small x$ .

Letting $\small u=lnx$ and $\small v=ln(lnx)$

(which means $\small u'=1/x$ and $\small v'=1/(xlnx)$ ) we get our derivative to be $\small ln(lnx)/x+1/x$ .

Now we have $\small y'/y=ln(lnx)/x+1/x$ , but $\small y=lnx^{lnx}$ , so subbing that in we get

$\small \small y'/(lnx^{lnx})=ln(lnx)/x+1/x$ .

Multiplying both sides by $\small lnx^{lnx}$ , we get

$\small y'=lnx^{lnx}(ln(lnx)/x+1/x)$ .

That is our derivative.

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Example Question #1 : First And Second Derivatives Of Functions

The position of a car is given by the following function:

$f(x)=\cos(x)e^{11x}+\csc(x)$

What is the velocity function of the car?

Possible Answers:

$-\sin(x)e^{11x}+11\cos(x)e^{11x}+\csc(x)\cot(x)$

$-\sin(x)e^{11x}+11\cos(x)e^{11x}-\csc(x)\cot(x)$

$-\sin(x)e^{11x}+11\cos(x)e^{11x}$

$-\sin(x)e^{11x}+\cos(x)e^{11x}-\csc(x)\cot(x)$

$\sin(x)e^{11x}+11\cos(x)e^{11x}-\csc(x)\cot(x)$

Correct answer:

$-\sin(x)e^{11x}+11\cos(x)e^{11x}-\csc(x)\cot(x)$

Explanation:

The velocity function of the car is equal to the first derivative of the position function of the car, and is equal to

$-\sin(x)e^{11x}+11\cos(x)e^{11x}-\csc(x)\cot(x)$

The derivative was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u \frac{\mathrm{du} }{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \csc(x)=-\csc(x)\cot(x)$

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Example Question #211 : Derivative Review

Find the derivative of the following function:

$f(x)=\sec(3x)+e^{2x^2+1}$

Possible Answers:

$3\sec(3x)\tan(3x)+4xe^{2x^2+1}$

$3\sec(3x)\tan(3x)-e^{2x^2+1}$

$\sec(3x)\tan(3x)+4xe^{2x^2+1}$

$3\sec(x)\tan(x)+4xe^{2x^2+1}$

$3\sec(3x)\tan(3x)+e^{2x^2+1}$

Correct answer:

$3\sec(3x)\tan(3x)+4xe^{2x^2+1}$

Explanation:

The derivative of the function is equal to

$f'(x)=3\sec(3x)\tan(3x)+4xe^{2x^2+1}$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{du} }{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \sec(x)=\sec(x)\tan(x)$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #258 : Acceleration

Example Question #291 : How To Find Acceleration

Example Question #211 : Calculus 3

Example Question #1 : Computation Of Derivatives

Example Question #213 : Calculus 3

Example Question #211 : Calculus 3

Example Question #1 : First And Second Derivatives Of Functions

Example Question #214 : Calculus 3

Example Question #1 : First And Second Derivatives Of Functions

Example Question #211 : Derivative Review

All Calculus 3 Resources

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