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Calculus 2 : First and Second Derivatives of Functions

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Example Questions

Example Question #71 : First And Second Derivatives Of Functions

Find the first derivative of the following function:

$f(x)=e^{\sin^2(x)}$

Possible Answers:

$-2\sin(x)\cos(x)e^{\sin^2(x)}$

$\sin(x)\cos(x)e^{\sin^2(x)}$

$e^{\sin^2(x)}$

$2\sin(x)\cos(x)e^{\sin^2(x)}$

Correct answer:

$2\sin(x)\cos(x)e^{\sin^2(x)}$

Explanation:

The first derivative of the function is equal to

$f'(x)=2\sin(x)\cos(x)e^{\sin^2(x)}$

and was found using the following rules:

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

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Example Question #271 : Derivative Review

Find the first derivative of the function:

$g(x)=\frac{x^2}{e^{3x}}$

Possible Answers:

$\frac{2xe^{3x}-3x^2e^{3x}}{e^{6x}}$

$\frac{2xe^{3x}-3x^2e^{3x}}{e^{3x^2}}$

$\frac{2xe^{3x}-x^2e^{3x}}{e^{6x}}$

$\frac{2xe^{3x}+3x^2e^{3x}}{e^{6x}}$

Correct answer:

$\frac{2xe^{3x}-3x^2e^{3x}}{e^{6x}}$

Explanation:

The first derivative of the function is equal to

$g'(x)=\frac{2xe^{3x}-3x^2e^{3x}}{e^{6x}}$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} \frac{f(x)}{g(x)}=\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{du} }{\mathrm{d} x}$

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Example Question #272 : Derivative Review

Find the first derivative of the function:

$f(x)=e^{3x^4}\cos(x)$

Possible Answers:

$16x^3e^{3x^4}\cos(x)-e^{3x^4}\sin(x)$

$12x^3e^{3x^4}\cos(x)+e^{3x^4}\sin(x)$

$12x^3e^{3x^4}\sin(x)-e^{3x^4}\cos(x)$

$12x^3e^{3x^4}\cos(x)-e^{3x^4}\sin(x)$

Correct answer:

$12x^3e^{3x^4}\cos(x)-e^{3x^4}\sin(x)$

Explanation:

The derivative of the function is equal to

$f'(x)=12x^3e^{3x^4}\cos(x)-e^{3x^4}\sin(x)$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{du} }{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$

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Example Question #273 : Derivative Review

Find the derivative of the function:

$f(x)=\frac{x^2+2x+3}{xe^x}$

Possible Answers:

$\frac{xe^x(2x+2)-(x^2+2x+3)(e^x+xe^x)}{(xe^x)^2}$

$\frac{xe^x(2x+2)-(x^2+2x+3)(e^x+xe^x)}{xe^x}$

$\frac{(2x+2)-(x^2+2x+3)}{x}$

$\frac{xe^x(2x+2)+(x^2+2x+3)(e^x+xe^x)}{(xe^x)^2}$

$\frac{(2x+2)-(x^2+2x+3)(e^x)}{xe^x}$

Correct answer:

$\frac{xe^x(2x+2)-(x^2+2x+3)(e^x+xe^x)}{(xe^x)^2}$

Explanation:

The derivative of the function is equal to

$f'(x)=\frac{xe^x(2x+2)-(x^2+2x+3)(e^x+xe^x)}{(xe^x)^2}$

and was found using the following rules:

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Example Question #71 : First And Second Derivatives Of Functions

What is the derivative of $f(x)=\frac{1}{2}x^2+6x-1?$

Possible Answers:

f{}'(x)=x-6

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

f{}'(x)=2x+6

f{}'(x)=x+6

Correct answer:

f{}'(x)=x+6

Explanation:

To take the derivative, remember to multiply the exponent by the coefficient in front of the x term and then subtract one from the exponent. Therefore, your answer should be: $f{}'(x)=(\frac{1}{2}(2))x+6=x+6$ .

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Example Question #73 : First And Second Derivatives Of Functions

What is the acceleration function if the position function is x(t)=4t^2-3t+1 ?

Possible Answers:

a(t)=.8t

$a(t)=\frac{1}{8}$

a(t)=8t-1

a(t)=8t

a(t)=8

Correct answer:

a(t)=8

Explanation:

Recall that the acceleration function is the second derivative of the position function. So, the first step is taking the first derivative. Remember to multiply the exponent by the coefficient in front of the x term and then subtract the exponent by 1. The first derivative is v(t)=8t-3 . Then, take the derivative of the velocity function to get the acceleration function, which is a(t)=8 .

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Example Question #72 : First And Second Derivatives Of Functions

What is the derivative of f(x)=-2x^2+x-3 ?

Possible Answers:

f{}'(x)=-4x+1

f{}'(x)=4x+1

f{}'(x)=-x+1

f{}'(x)=-4x-1

f{}'(x)=-4x

Correct answer:

f{}'(x)=-4x+1

Explanation:

Remember that when taking the derivative, multiply the exponent by the coefficient in front of the x and then subtract one from the exponent. Therefore, your answer is: f{}'(x)=-4x+1 .

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Example Question #1403 : Calculus Ii

Find the derivative of x^2y-2x+y=3x .

Possible Answers:

$\frac{dy}{dx}=\frac{5-2xy}{x^2+1}$

$\frac{dy}{dx}=\frac{5+2xy}{x^2+1}$

$\frac{dy}{dx}=\frac{5-xy}{x^2+1}$

$\frac{dy}{dx}=\frac{5-2xy}{x^2-1}$

$\frac{dy}{dx}=\frac{5-2xy}{x+1}$

Correct answer:

$\frac{dy}{dx}=\frac{5-2xy}{x^2+1}$

Explanation:

To find the derivative, remember to use implicit differentiation.

To find the derivative take the derivative of each term.

In this particular case the power rule,

$\frac{d}{dx}x^n=nx^{n-1}$

and the product rule,

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

will be applied to solve.

Your first step should look like this:

$x^2\frac{dy}{dx}+2xy-2+\frac{dy}{dx}=3$ .

Then, solve for $\frac{dy}{dx}$ .

The next step should look like:

$\frac{dy}{dx}(x^2+1)=3+2-2xy$ .

Thus, your answer is:

$\frac{dy}{dx}=\frac{5-2xy}{x^2+1}$ .

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Example Question #1404 : Calculus Ii

What is the derivative of $f(x)=\frac{1}{4}x^{\frac{3}{2}}$ ?

Possible Answers:

$f{}'(x)=\sqrt{x}$

$f{}'(x)=\frac{3}{8}\sqrt{x^3}$

$f{}'(x)=\frac{3}{8}\sqrt{x}$

$f{}'(x)=\frac{3}{4}\sqrt{x}$

$f{}'(x)=\frac{1}{8}\sqrt{x}$

Correct answer:

$f{}'(x)=\frac{3}{8}\sqrt{x}$

Explanation:

To find the derivative, multiply the exponent by the coefficient in front of the x and then subtract 1 from the exponent.

Therefore, the first step is:

$\frac{1}{4}(\frac{3}{2})(x^{\frac{1}{2}})$ .

Multiply like terms to get your answer of

$f{}'(x)=\frac{3}{8}\sqrt{x}$ .

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Example Question #73 : First And Second Derivatives Of Functions

What is the derivative of f(x)=16x^2-3x+1 ?

Possible Answers:

f{}'(x)=32x-3

f{}'(x)=32x+3

f{}'(x)=2x-3

f{}'(x)=16x-3

f{}'(x)=32x

Correct answer:

f{}'(x)=32x-3

Explanation:

To find the derivative, remember to multiply the exponent by the coefficient in front of the x and then subtract 1 from the exponent.

Therefore, the derivative is:

f{}'(x)=32x-3.

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Calculus 2 : First and Second Derivatives of Functions

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #71 : First And Second Derivatives Of Functions

Example Question #271 : Derivative Review

Example Question #272 : Derivative Review

Example Question #273 : Derivative Review

Example Question #71 : First And Second Derivatives Of Functions

Example Question #73 : First And Second Derivatives Of Functions

Example Question #72 : First And Second Derivatives Of Functions

Example Question #1403 : Calculus Ii

Example Question #1404 : Calculus Ii

Example Question #73 : First And Second Derivatives Of Functions

All Calculus 2 Resources

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