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AP Calculus AB : AP Calculus AB

Study concepts, example questions & explanations for AP Calculus AB

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3 Diagnostic Tests 164 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

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Example Question #91 : Ap Calculus Ab

Find the limit of the function below using L'Hopital's rule

$\binom{lim}{q\rightarrow 4}\frac{q^2-2q-8}{sin(q\pi)}$

Possible Answers:

$\binom{lim}{q\rightarrow 4}\frac{q^2-2q-8}{sin(q\pi)}=\frac{0}{0}$

$\binom{lim}{q\rightarrow 4}\frac{q^2-2q-8}{sin(q\pi)}=\frac{6}{\pi}$

$\binom{lim}{q\rightarrow 4}\frac{q^2-2q-8}{sin(q\pi)}=6\pi$

$\binom{lim}{q\rightarrow 4}\frac{q^2-2q-8}{sin(q\pi)}=-6\pi$

$\binom{lim}{q\rightarrow 4}\frac{q^2-2q-8}{sin(q\pi)}=\frac{\pi}{6}$

Correct answer:

$\binom{lim}{q\rightarrow 4}\frac{q^2-2q-8}{sin(q\pi)}=\frac{6}{\pi}$

Explanation:

*****************************************************************

STEPS

*****************************************************************

Given:

$\binom{lim}{q\rightarrow 4}\frac{q^2-2q-8}{sin(q\pi)}$

Plug in 4 for q to test to use L'Hopital's rule

$\binom{lim}{q\rightarrow 4}=\frac{4^2-2(4)-8}{sin(4\pi)}$

$\binom{lim}{q\rightarrow 4}=\frac{0}{0}$

From this indeterminate form, realize we can now use L'Hopital's rule, deriving the expressions in the numerator and denominator independently of one another

$\binom{lim}{q\rightarrow 4}\frac{d(q^2-2q-8)}{d(sin(q\pi))}$

$\binom{lim}{q\rightarrow 4}\frac{2q-2}{\pi(cos(q\pi))}$ (pi comes from chain rule with qpi)

$\binom{lim}{q\rightarrow 4}=\frac{2(4)-2}{\pi(cos((4)\pi))}$

$\binom{lim}{q\rightarrow 4}=\frac{6}{\pi*1}$

Thus, we arrive at our correct answer:

*****************************************************************

CORRECT ANSWER

*****************************************************************

${\color{Red} \binom{lim}{q\rightarrow 4}\frac{q^2-2q-8}{sin(q\pi)}=\frac{6}{\pi}}$

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Example Question #71 : Computation Of The Derivative

Find the derivative of the function:

$s(r)=7^{2r}$

Possible Answers:

$s'(r)=7^{2r}*2ln(7)$

s'(r)=rln(7)

s'(r)=-2r*2ln(7)

s'(r)=2r*2ln(7)

$s'(r)=7^{2r}*7ln(2)$

Correct answer:

$s'(r)=7^{2r}*2ln(7)$

Explanation:

*****************************************************************

STEPS

*****************************************************************

Given:

$s(r)=7^{2r}$

Understand the derivative of a^x:

d(a^x)=a^xln(a)

Thus:

$s'(r)=7^{2r}*ln(7)*(chain)$

We find the chain by deriving the compound operation "2r"

d(2r)=2

Plug 2 in for the chain:

$s'(r)=7^{2r}*ln(7)*2$

Thus, we arrive at the correct answer:

*****************************************************************

CORRECT ANSWER

*****************************************************************

$s'(r)=7^{2r}*2ln(7)$

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Example Question #71 : Derivatives Of Functions

Find the derivative of the function:

$y=cos^{-1}(x^2)$

Possible Answers:

$y'=\frac{\sqrt{1-x^4}}{2x}$

$y'=-\frac{\sqrt{1-x^2}}{x}$

$y'=-\frac{\sqrt{1-x^4}}{2x}$

$y'=\frac{-2x}{\sqrt{1-x^4}}$

$y'=\frac{2x}{\sqrt{1-x^4}}$

Correct answer:

$y'=\frac{-2x}{\sqrt{1-x^4}}$

Explanation:

*****************************************************************

STEPS

*****************************************************************

Given:

$y=cos^{-1}(x^2)$

$y'=d(cos^{-1}(x^2))*(chain)$

$d(cos^-1(x^2))=\frac{-1}{\sqrt{1-(x^2)^2}}$

$d(cos^-1(x^2))=\frac{-1}{\sqrt{1-x^4}}$

Now, take the derivative of x^2, and plug in for the "chain"

$d(cos^-1(x^2))=\frac{-1}{\sqrt{1-x^4}}*(chain)$

$d(cos^-1(x^2))=\frac{-1}{\sqrt{1-x^4}}*(d(x^2))$

$d(cos^-1(x^2))=\frac{-1}{\sqrt{1-x^4}}*2x$

Multiplying, we arrive at the correct answer

*****************************************************************

CORRECT ANSWER

*****************************************************************

$d(cos^-1(x^2))={\color{Red} y'=\frac{-2x}{\sqrt{1-x^4}}}$

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Example Question #94 : Ap Calculus Ab

Find the limit of the function below using L'Hopital's Rule

$\binom{lim}{y\rightarrow 2}\frac{y^3-7y^2+10y}{y^2+y-6}$

Possible Answers:

The limit does not exist

$\binom{lim}{y\rightarrow 2}\frac{y^3-7y^2+10y}{y^2+y-6}=\frac{6}{5}$

$\binom{lim}{y\rightarrow 2}\frac{y^3-7y^2+10y}{y^2+y-6}=\frac{-6}{5}$

$\binom{lim}{y\rightarrow 2}\frac{y^3-7y^2+10y}{y^2+y-6}=\frac{-5}{6}$

$\binom{lim}{y\rightarrow 2}\frac{y^3-7y^2+10y}{y^2+y-6}=\frac{5}{6}$

Correct answer:

$\binom{lim}{y\rightarrow 2}\frac{y^3-7y^2+10y}{y^2+y-6}=\frac{-6}{5}$

Explanation:

*****************************************************************

STEPS

*****************************************************************

Given:

$\binom{lim}{y\rightarrow 2}\frac{y^3-7y^2+10y}{y^2+y-6}$

Try the limit. Plug in two for y and check the result:

$\binom{lim}{y\rightarrow 2}=\frac{(2)^3-7(2)^2+10(2)}{(2)^2+(2)-6}$

$\binom{lim}{y\rightarrow 2}=\frac{8-28+20}{4+2-6}$

$\binom{lim}{y\rightarrow 2}=\frac{0}{0}$

Thus, we realize me must use L'Hopital's Rule on the original quotient, deriving the expressions in the numerator and denominator independently

$\binom{lim}{y\rightarrow 2}\frac{y^3-7y^2+10y}{y^2+y-6}$

$\binom{lim}{y\rightarrow 2}\frac{d(y^3-7y^2+10y)}{d(y^2+y-6)}$

$\binom{lim}{y\rightarrow 2}\frac{3y^2-14y+10}{2y+1}$

Try the limit once more:

$\binom{lim}{y\rightarrow 2}=\frac{3(2)^2-14(2)+10}{2(2)+1}$

$\binom{lim}{y\rightarrow 2}=\frac{12-28+10}{4+1}$

$\binom{lim}{y\rightarrow 2}=\frac{-16+10}{5}$

Simplifying the numerator, we arrive at the correct answer:

*****************************************************************

CORRECT ANSWER

*****************************************************************

${\color{Red} \binom{lim}{y\rightarrow 2}\frac{y^3-7y^2+10y}{y^2+y-6}=\frac{-6}{5}}$

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Example Question #95 : Ap Calculus Ab

Find the derivative of the function

$g(x)=e^{\lambda x^3}+\cos(\lambda x)$ ,

where $\lambda$ is a constant.

Possible Answers:

$-\lambda\sin(\lambda x)$

$3 \lambda x^2 e^{\lambda x^3}-\lambda\sin(\lambda x)$

$3 \lambda x^2 e^{\lambda x^3}-\sin(\lambda x)$

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

Correct answer:

$3 \lambda x^2 e^{\lambda x^3}-\lambda\sin(\lambda x)$

Explanation:

The derivative of the function is equal to

$g'(x)=3 \lambda x^2 e^{\lambda x^3}-\lambda\sin(\lambda x)$

and was found using the following rules:

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

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Example Question #96 : Ap Calculus Ab

Find the derivative of the function:

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

The derivative of the function is equal to

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

and was found using the following rules:

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

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Example Question #97 : Ap Calculus Ab

Find the derivative of the function:

$f=\ln(6x+x^2)$

Possible Answers:

$-\frac{2x+6}{x^2+6x}$

$\frac{1}{x^2+6x}$

$\frac{8}{x+6}$

$\frac{2x+6}{x^2+6x}$

Correct answer:

$\frac{2x+6}{x^2+6x}$

Explanation:

The derivative of the function is equal to

$f'=\frac{2x+6}{x^2+6x}$

and was found using the following rules:

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

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Example Question #98 : Ap Calculus Ab

Find the derivative of the following function:

$f(x)=\frac{x^2+3x+1}{x^4}$

Possible Answers:

$\frac{2x^2+9x+4}{x^5}$

$\frac{6x^2+15x+4}{x^5}$

$\frac{-2x^2-9x-4}{x^5}$

Correct answer:

$\frac{-2x^2-9x-4}{x^5}$

Explanation:

The derivative of the function is equal to

$f'(x)=\frac{-2x^2-9x-4}{x^5}$

and was found using the following rules:

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

The derivative was simplified from

$\frac{-2x^5-9x^4-4x^3}{x^8}$

to its most simple form.

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Example Question #99 : Ap Calculus Ab

Find the derivative of the function:

$f(x)=\frac{x\cos(x)}{x+3}$

Possible Answers:

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

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

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

$\frac{\cos(x)-x\sin(x)-x\cos(x)}{x+3}$

Correct answer:

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

Explanation:

The derivative of the function is equal to

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

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} \frac{f(x)}{g(x)}=\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \sin(x)=\cos(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

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Example Question #100 : Ap Calculus Ab

Find the derivative of the function:

$f=2\sec(x)+2\csc(x)$

Possible Answers:

$2\sec(x)\tan(x)-2\csc(x)\cot(x)$

$2\csc(x)\cot(x)-2\sec(x)\tan(x)$

$2\ln\left |\sec(x)+\tan(x) \right |-2\ln\left |\csc(x)+\cot(x) \right |$

$-2\sec(x)\tan(x)-2\csc(x)\cot(x)$

Correct answer:

$2\sec(x)\tan(x)-2\csc(x)\cot(x)$

Explanation:

The derivative of the function is equal to

$f'=2\sec(x)\tan(x)-2\csc(x)\cot(x)$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} \sec(x)=\sec(x)\tan(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x }\csc(x)=-\csc(x)\cot(x)$

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AP Calculus AB : AP Calculus AB

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #91 : Ap Calculus Ab

Example Question #71 : Computation Of The Derivative

Example Question #71 : Derivatives Of Functions

Example Question #94 : Ap Calculus Ab

Example Question #95 : Ap Calculus Ab

Example Question #96 : Ap Calculus Ab

Example Question #97 : Ap Calculus Ab

Example Question #98 : Ap Calculus Ab

Example Question #99 : Ap Calculus Ab

Example Question #100 : Ap Calculus Ab

All AP Calculus AB Resources

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