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AP Calculus AB : Derivatives

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Example Questions

Example Question #61 : Derivatives Of Functions

Find the derivative of the function

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

Possible Answers:

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

$f'=\cos(x^2)-x\sin(x^2)$

$f'=-\cos(x^2)-2x^2\sin(x^2)$

$f'=\cos(x^2)-2x^2\sin(x^2)$

Correct answer:

$f'=\cos(x^2)-2x^2\sin(x^2)$

Explanation:

To find the derivative of the function, we use both the product rule and the chain rule

$f'=x'\cos(x^2)+x\cos(x^2)'=\cos(x^2)-x\sin(x^2)(2x)=\cos(x^2)-2x^2\sin(x^2)$

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Example Question #62 : Derivatives Of Functions

Find the derivative of the function $f=\frac{x^5}{\sin(2x)}$

Possible Answers:

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

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

$f'=\frac{5x^4\sin(2x)+2x^5\cos(2x)}{\sin^2(2x)}$

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

Correct answer:

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

Explanation:

To find the derivative of the function, we use both the quotient rule and the chain rule

$f'=(x^5'*\sin(2x)-x^5\sin(2x)')/\sin^2(2x))$

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

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Example Question #63 : Derivatives Of Functions

If $f(x)=\sin x$ then $f'\left({\frac{\pi }{4}}\right)=$

Possible Answers:

$\frac{{\sqrt{3}}}{2}$

$\frac{\sqrt{2}}{2}$

Correct answer:

$\frac{\sqrt{2}}{2}$

Explanation:

To calculate the derivative of this function at the desired point, first recall that,

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

$f(x)=\sin x$

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

Now, substitute the value into the derivative function to solve.

$f'(\frac{\pi }{4}) = cos\frac{\pi }{4} = cos(45^{\circ}) = \frac{\sqrt{2}}{2}$

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Example Question #64 : Derivatives Of Functions

Let $f(x)=x^2-\frac{1}{1-x^2}$ . Which of the following gives the equation of the line normal to $f(x)$ when x=2 ?

Possible Answers:

$27x+111y=535$

$27x + 96y = 470$

$32x+9y=103$

$192x + 54y = 735$

$32x-9y=25$

Correct answer:

$27x + 96y = 470$

Explanation:

We are asked to find the normal line. This means we need to find the line that is perpendicular to the tangent line at x=2 . In order to find the tangent line, we will need to evaluate the derivative of f(x) at x=2 .

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

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

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

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

$f'(2)=4-4(\frac{1}{9})=\frac{32}{9}$

The slope of the tangent line at x=2 is $\frac{32}{9}$ . Because the tangent line and the normal line are perpendicular, the product of their slopes must equal .

(slope of tangent)(slope of normal) =

$\left ( \frac{32}{9} \right )\cdot \left ( slope\ of\ normal \right )=-1$

$slope\ of\ normal=-\frac{9}{32}$

We now have the slope of the normal line. Once we find a point through which it passes, we will have enough information to derive its equation.

Since the normal line passes through the function at x=2 , it will pass through the point $\left ( 2,f(2) \right )$ . Be careful to use the original equation for , not its derivative.

$f(2)=2^2-(1-4)^{-1}=4-(-\frac{1}{3})=\frac{13}{3}$

The normal line has a slope of $-\frac{9}{32}$ and passes through the piont $\left ( 2,\frac{13}{3} \right )$ . We can now use point-slope form to find the equation of the normal line.

$y-\frac{13}{3}=-\frac{9}{32}(x-2)$

Multiply both sides by 32(3)=96 .

$96y-416=-27(x-2)$

$27x + 96y = 470$

The answer is $27x + 96y = 470$ .

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Example Question #65 : Derivatives Of Functions

If the position of a particle over time is represented by $p=t^{3}-16t^{2}-7t$ then what is the particle's instantaneous acceleration at t=4 ?

Possible Answers:

Correct answer:

Explanation:

The answer is .

Since velocity is the first derivative of the position function, take the derivative once. Then, recall that the acceleration function is the second derivative of position thus the derivative needs to be taken one more time.

$p=t^{3}-16t^{2}-7t$

$velocity =p'=3t^{2}-32t-7$

$acceleration =p''=6t-32$

$p''(4)=6(4)-32 = -8$

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Example Question #66 : Derivatives Of Functions

Consider the function:

$f(x) = -6e^{-x} + x$

The relative minimum for this function is at:

Possible Answers:

There is no relative minimum, because the first derivative is always positive.

$x = \frac{-1}{6}$

None of the other answers.

x = 0

x = 10

Correct answer:

There is no relative minimum, because the first derivative is always positive.

Explanation:

To find any relative minimum, one first needs to find the critical points by setting the first derivative equal to zero:

$f'(x) = 6e^{-x} + 1 = 0$

However, the first derivative is positive for all real values of x, since the exponential function is always positive. Thus, there are no values for which f'(x) = 0 , and therefore no critical points and no relative minimum.

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Example Question #71 : Derivatives Of Functions

Find the derivative of the function:

g(x)=sinx*tanx

Possible Answers:

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

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

None of the other derivatives are correct

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

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

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

(Note: If you require or desire further explanation for the assumed knowledge, ask your Varsity Tutor for clarity on the concepts!)

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Example Question #88 : Ap Calculus Ab

Find the first derivative of the function:

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

Possible Answers:

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

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

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

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

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

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 #72 : Derivatives Of Functions

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:

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

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

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

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

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

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 #73 : Derivatives Of Functions

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

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

Possible Answers:

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

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

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

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

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

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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AP Calculus AB : Derivatives

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #61 : Derivatives Of Functions

Example Question #62 : Derivatives Of Functions

Example Question #63 : Derivatives Of Functions

Example Question #64 : Derivatives Of Functions

Example Question #65 : Derivatives Of Functions

Example Question #66 : Derivatives Of Functions

Example Question #71 : Derivatives Of Functions

Example Question #88 : Ap Calculus Ab

Example Question #72 : Derivatives Of Functions

Example Question #73 : Derivatives Of Functions

All AP Calculus AB Resources

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