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AP Calculus AB : Computation of the Derivative

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Example Question #87 : Ap Calculus Ab

Find the derivative of the function:

g(x)=sinx*tanx

Possible Answers:

g'(x)=2sinxsec^2x+sinx

g'(x)=-2sinxsec^2x+sinx

g'(x)=sinxsec^2x+sinx

g'(x)=-sinxsec^2x+sinx

None of the other derivatives are correct

Correct answer:

g'(x)=sinxsec^2x+sinx

Explanation:

We are given the function:

g(x)=sinx*tanx

To take the derivative of this, we must understand 3 things:

The product rule for derivatives
The derivative of the sin function
The derivative of the tan function

We can apply the product rule as follows:

g'(x)=[sinx*d(tanx)]+[tanx*d(sinx)]

g'(x)=[sinx*sec^2x]+[tanx*cosx]

Simplifying once more, the answer becomes:

g'(x)=sinx*sec^2x+sinx

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Example Question #88 : Ap Calculus Ab

Find the first derivative of the function:

$r=sec(\theta)*csc(\theta)$

Possible Answers:

$r'=\frac{(sec^2(\theta)-csc^2(\theta))}{2}$

$r'=sec^2(\theta)+csc^2(\theta)$

$r'=2(sec^2(\theta)-csc^2(\theta))$

$r'=-2(sec^2(\theta)-csc^2(\theta))$

$r'=sec^2(\theta)-csc^2(\theta)$

Correct answer:

$r'=sec^2(\theta)-csc^2(\theta)$

Explanation:

We are given the function:

$r=sec(\theta)*csc(\theta)$

Our first step in tackling this problem is to apply the product rule for derivatives:

$r'=[sec(\theta)*d(csc(\theta))]+[csc(\theta)*d(sec(\theta))]$

$r'=[sec(\theta)*(-csc(\theta)cot(\theta))]+[csc(\theta)*(sec(\theta)tan(\theta))]$

To simplify the above further, it is best to write everything in terms of sin and cos, like so:

$r'=(\frac{1}{cos(\theta)}*\frac{-1}{sin(\theta)}*\frac{cos(\theta)}{sin(\theta)})+(\frac{1}{sin(\theta)}*\frac{1}{cos(\theta)}*\frac{sin(\theta)}{cos(\theta)})$

We can simplify both sides by multiplying the components together and canceling out like so:

$r'=(\frac{1}{cos(\theta)}*\frac{-1}{sin(\theta)}*\frac{cos(\theta)}{sin(\theta)})+(\frac{1}{sin(\theta)}*\frac{1}{cos(\theta)}*\frac{sin(\theta)}{cos(\theta)})$

$r'=\frac{-1}{sin^2(\theta)}+\frac{1}{cos^2(\theta)}$

$r'=\frac{1}{cos^2(\theta)}-\frac{1}{sin^2(\theta)}$

Simplifying our last terms, we arrive at the correct answer:

$r'=sec^2(\theta)-csc^2(\theta)$

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Example Question #89 : Ap Calculus Ab

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

$\binom{lim}{y\rightarrow 2}\frac{y^2+4y-12}{y-2}$

Possible Answers:

$\binom{lim}{y\rightarrow 2}=8$

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

$\binom{lim}{y\rightarrow 2}=\infty$

$\binom{lim}{y\rightarrow 2}=-\infty$

$\binom{lim}{y\rightarrow 2}=-8$

Correct answer:

$\binom{lim}{y\rightarrow 2}=8$

Explanation:

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

STEPS

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

First, try to plug in the 2, and check the result.

When doing this, the limit becomes an indeterminate form of

$\frac{0}{0}$

Thus, we continue with L'Hôpital's rule, deriving the expressions in the numerator and denominator like so:

$\binom{lim}{y\rightarrow 2}\frac{y^2+4y-12}{y-2}$

$\binom{lim}{y\rightarrow 2}\frac{d(y^2+4y-12)}{d(y-2)}$

$\binom{lim}{y\rightarrow 2}\frac{2y+4}{1}$

Now, plug 2 in for y, to check again

$\binom{lim}{y\rightarrow 2}=\frac{2(2)+4}{1}$

$\binom{lim}{y\rightarrow 2}=\frac{8}{1}$

Thus, our correct answer is:

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

CORRECT ANSWER

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

${\color{Red} \binom{lim}{y\rightarrow 2}=8}$

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Example Question #90 : Ap Calculus Ab

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

$\binom{lim}{s\rightarrow \pi}\frac{sin(s)}{sin(3s)}$

Possible Answers:

$\binom{lim}{s\rightarrow \pi}\frac{sin(s)}{sin(3s)}=3$

$\binom{lim}{s\rightarrow \pi}\frac{sin(s)}{sin(3s)}=\frac{1}{3}$

$\binom{lim}{s\rightarrow \pi}\frac{sin(s)}{sin(3s)} =0$

$\binom{lim}{s\rightarrow \pi}\frac{sin(s)}{sin(3s)}=-3$

$\binom{lim}{s\rightarrow \pi}\frac{sin(s)}{sin(3s)}=-\frac{1}{3}$

Correct answer:

$\binom{lim}{s\rightarrow \pi}\frac{sin(s)}{sin(3s)}=\frac{1}{3}$

Explanation:

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

STEPS

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

We are asked to find the limit:

$\binom{lim}{s\rightarrow \pi}\frac{sin(s)}{sin(3s)}$

First, plug pi in for s to check if we can use L'Hopital's Rule

$\binom{lim}{s\rightarrow \pi}=\frac{sin(\pi)}{sin(3\pi)}$

$\binom{lim}{s\rightarrow \pi}=\frac{0}{0}$

We have the indeterminate form zero over zero, so we continue with L'Hopital's Rule, deriving the expressions in the numerator and denominator, independently of each other:

$\binom{lim}{s\rightarrow \pi}\frac{sin(s)}{sin(3s)}$

$\binom{lim}{s\rightarrow \pi}\frac{cos(s)}{3cos(3s)}$ (The 3 before the cos is from the derivative of 3s (=3)

Now, plug pi in once more to check:

$\binom{lim}{s\rightarrow \pi}=\frac{cos(\pi)}{3cos(\pi)}$

$\binom{lim}{s\rightarrow \pi}=\frac{-1}{-3}$

Thus, canceling the negatives, we arrive at the correct answer:

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

CORRECT ANSWER

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

${\color{Red} \binom{lim}{s\rightarrow \pi}=\frac{1}{3}}$

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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 #93 : Ap Calculus Ab

Find the derivative of the function:

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

Possible Answers:

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

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

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

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

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

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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AP Calculus AB : Computation of the Derivative

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #87 : Ap Calculus Ab

Example Question #88 : Ap Calculus Ab

Example Question #89 : Ap Calculus Ab

Example Question #90 : Ap Calculus Ab

Example Question #91 : Ap Calculus Ab

Example Question #71 : Computation Of The Derivative

Example Question #93 : Ap Calculus Ab

Example Question #94 : Ap Calculus Ab

Example Question #95 : Ap Calculus Ab

Example Question #96 : Ap Calculus Ab

All AP Calculus AB Resources

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