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Calculus 2 : First and Second Derivatives of Functions

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Example Questions

Example Question #311 : Derivative Review

What is the derivative of $f(x)=5x^{\frac{4}{5}}-2x$ ?

Possible Answers:

$f{}'(x)=\frac{4}{x^{\frac{1}{5}}}-2x$

$f{}'(x)=\frac{2}{x^{\frac{1}{5}}}-2$

$f{}'(x)=\frac{4}{x^{\frac{1}{5}}}-2$

$f{}'(x)=\frac{4}{x^-{\frac{1}{5}}}-2$

$f{}'(x)=\frac{4}{x^{\frac{1}{5}}}$

Correct answer:

$f{}'(x)=\frac{4}{x^{\frac{1}{5}}}-2$

Explanation:

Remember that when taking the derivative, multiply the exponent by the coefficient in front of the x term and then subtract one from the exponent: $5(\frac{4}{5})x^{\frac{-1}{5}}-2$ . Simplify and make sure you don't have a negative exponent (to make it positive, put it in the denominator). Therefore, your answer is $f{}'(x)=\frac{4}{x^{\frac{1}{5}}}-2$ .

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Example Question #112 : First And Second Derivatives Of Functions

What is the derivative of $f(x)=9x^{\frac{5}{4}}-3x?$

Possible Answers:

$f{}'(x)=\frac{45}{4}x^{\frac{1}{4}}-3$

$f{}'(x)=\frac{45}{2}x^{\frac{1}{4}}-3$

$f{}'(x)=\frac{45}{4}x^{\frac{1}{4}}-3x$

$f{}'(x)=\frac{45}{4}x^{\frac{3}{4}}-3$

$f{}'(x)=\frac{45}{4}x^{\frac{1}{4}}-5$

Correct answer:

$f{}'(x)=\frac{45}{4}x^{\frac{1}{4}}-3$

Explanation:

Remember that when taking the derivative, multiply the exponent by the coefficient in front of the x term and then subtract one from the exponent.

Therefore, your answer should look like this:

$f{}'(x)=\frac{45}{4}x^{\frac{1}{4}}-3$

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Example Question #111 : First And Second Derivatives Of Functions

What is the derivative of f(x)=3x^3-2x^2+4x-1?

Possible Answers:

f{}'(x)=9x+4

f{}'(x)=18x-4

$f{}'(x)=9x^2-4x+4$

$f{}'(x)=9x^2-4x$

Correct answer:

$f{}'(x)=9x^2-4x+4$

Explanation:

Recall that when taking the derivative, multiply the exponent by the coefficient in front of the x term and then subtract one from the exponent.

Therefore, after differentiating, you get

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

$f{}'(x)=9x^2-4x+4$ .

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Example Question #114 : First And Second Derivatives Of Functions

Find the derivative of the function:

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

Possible Answers:

$3x^2+e^x(\sin(x)-\cos(x)+4)$

$3x^2+e^x(\sin(x)+4)$

$3x^2+e^x(\cos(x)+4)$

$3x^2+e^x(\sin(x)+\cos(x))+e^{4x}$

$3x^2+e^x(\sin(x)+\cos(x)+4)$

Correct answer:

$3x^2+e^x(\sin(x)+\cos(x)+4)$

Explanation:

The derivative of the function is equal to

$f'(x)=3x^2+e^x(\sin(x)+\cos(x)+4)$

and was found using the following rules:

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

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Example Question #112 : First And Second Derivatives Of Functions

Find the second derivative of the function

$f(x)=x^2\sin(2x)$

Possible Answers:

$(2+4x)(\sin(2x))+8x(\cos(2x))$

$(2+4x^2)(\sin(2x))+8x(\cos(2x))$

$(2-4x^2)(\sin(2x))+8x\cos(2x)$

$(2-4x^2)(\sin(2x))+4x\cos(2x)$

Correct answer:

$(2-4x^2)(\sin(2x))+8x\cos(2x)$

Explanation:

To begin, you must take the first derivative of the function, which is:

$2x^2(\cos(2x))+2x(\sin(2x))$

Then, you take the derivative of the first derivative, which equals:

$4x(\cos(2x))-4x^2(\sin(2x))+2\sin(2x)+4x\cos(2x)$

This simplifies to:

$(2-4x^2)(\sin(2x))+8x(\cos(2x))$

The rules of differentiation that were used are:

$\frac{d}{dx}(f(x)g(x))=f'(x)g(x)+f(x)g'(x)$

$\frac{d}{dx}f(g(x))=f'(g(x))(g'(x))$

$\frac{d}{dx}(f(x)+g(x))=\frac{d}{dx}(f(x))+\frac{d}{dx}(g(x))$

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Example Question #312 : Derivatives

Find the derivative of the function:

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

Possible Answers:

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

$2xe^{x^2}\sec^2(e^{x^2})$

$2xe^{x^2}\sec(e^{x^2})$

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

Correct answer:

$2xe^{x^2}\sec^2(e^{x^2})$

Explanation:

The derivative of the function is equal to the following:

$f'(x)=2xe^{x^2}\sec^2(e^{x^2})$

and was found using the following rules:

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

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Example Question #313 : Derivatives

Find the first derivative of the function:

$f(x)=e^{\cos(x)}+e^{5x}$

Possible Answers:

$-\sin(x)e^{cos(x)}+5e^{5x}+C$

$-5\sin(x)e^{\cos(x)}$

$-\sin(x)e^{cos(x)}+5e^{5x}$

$\sin(x)e^{cos(x)}+5e^{5x}$

Correct answer:

$-\sin(x)e^{cos(x)}+5e^{5x}$

Explanation:

The first derivative of the function is equal to:

$f'(x)=-\sin(x)e^{cos(x)}+5e^{5x}$

and was found using the following rules:

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

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Example Question #313 : Derivative Review

Find the second derivative of the function:

$g(x)=x^{\frac{1}{2}}\cos(x)$

Possible Answers:

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

$\cos(x)(-\frac{1}{4}x^{-\frac{3}{2}}-x^{\frac{1}{2}})-\sin(x)(x^{\frac{1}{2}})$

$\frac{1}{2}x^{-\frac{1}{2}}\cos(x)-x^{\frac{1}{2}}\sin(x)$

$\cos(x)(-\frac{1}{4}x^{-\frac{3}{2}}-x^{\frac{1}{2}})+\sin(x)(x^{\frac{1}{2}})$

Correct answer:

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

Explanation:

The first derivative of the function is equal to

$g'(x)=\frac{1}{2}x^{-\frac{1}{2}}\cos(x)-x^{\frac{1}{2}}\sin(x)$

and was found using the following rules:

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

The second derivative of the function is equal to

$g''(x)=\cos(x)(-\frac{1}{4}x^{-\frac{3}{2}}-x^{\frac{1}{2}})$

and was found using the rules above, along with

$\frac{\mathrm{d} }{\mathrm{d} x} \sin(x)=\cos(x)$

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Example Question #314 : Derivatives

Find the derivative of the following function:

$f(x)=\ln(x^4e^x)$

Possible Answers:

$\frac{4}{x}+1$

$\frac{1}{x^4}$

$\frac{4}{x}$

$\frac{1}{x}+1$

$\frac{1}{x^4}+1$

Correct answer:

$\frac{4}{x}+1$

Explanation:

The derivative of the function is equal to

$f'(x)=\frac{4}{x}+1$

and was found using the following rules:

$\ln(ab)=\ln(a)+\ln(b)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \ln(u)=\frac{1}{u}\frac{\mathrm{du} }{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

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Example Question #117 : First And Second Derivatives Of Functions

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

Possible Answers:

$f{}'(x)=\frac{3}{x^{\frac{1}{4}}}$

$f{}'(x)=\frac{1}{x^{\frac{1}{4}}}-4$

$f{}'(x)=\frac{3}{x^{\frac{1}{4}}}+4$

$f{}'(x)=\frac{3}{x^{\frac{1}{4}}}-4$

$f{}'(x)=\frac{6}{x^{\frac{1}{4}}}-4$

Correct answer:

$f{}'(x)=\frac{3}{x^{\frac{1}{4}}}-4$

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:

$f{}'(x)=4(\frac{3}{4})x^{-\frac{1}{4}}-4$

Simplify and make the negative exponent positive by putting it on the denominator:

$f{}'(x)=\frac{3}{x^{\frac{1}{4}}}-4$ .

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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 #311 : Derivative Review

Example Question #112 : First And Second Derivatives Of Functions

Example Question #111 : First And Second Derivatives Of Functions

Example Question #114 : First And Second Derivatives Of Functions

Example Question #112 : First And Second Derivatives Of Functions

Example Question #312 : Derivatives

Example Question #313 : Derivatives

Example Question #313 : Derivative Review

Example Question #314 : Derivatives

Example Question #117 : First And Second Derivatives Of Functions

All Calculus 2 Resources

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