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AP Calculus AB : Derivatives

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Example Question #71 : Ap Calculus Ab

Use the method of your choice to find the derivative.

(x+5)(2x+8)

Possible Answers:

4x+18

2x+40

Correct answer:

4x+18

Explanation:

The easiest way to find this derivative is to FOIL, and then use the power rule.

2x^2+8x+10x+40=2x^2+18x+40

$\frac{d}{dx}2x^2+18x+40=4x+18$

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Example Question #72 : Ap Calculus Ab

Find the derivative.

$\sin (x)\cos (x)$

Possible Answers:

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

$-\cos (x)\sin (x)$

$\cos (x)\sin (x)$

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

Correct answer:

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

Explanation:

Use the product rule to find this derivative.

$\frac{d}{dx}\sin(x)\cos (x)=-\sin^2(x)+\cos^2(x)$

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Example Question #71 : Derivatives

Define $f (x) = \begin{Bmatrix} x^{2}+2x+ 3& x < 1 \\ a+ b \ln x & x \ge 1 \end{matrix}$

Evaluate and so that is both continuous and differentiable at x = 1 .

Possible Answers:

a= 4, b=4

a= 4, b=6

a= 6, b=2

a= 4, b=2

a= 6, b=4

Correct answer:

a= 6, b=4

Explanation:

For to be continuous at x = 1 , it must hold that

$\lim_{x \rightarrow 1^{-}} f(x) = \lim_{x \rightarrow 1^{+}} f(x)$ .

To find $\lim_{x \rightarrow 1^{-}} f(x)$ , we can use the definition of f(x) for all negative values of :

$\lim_{x \rightarrow 1^{-}} f(x) =\lim_{x \rightarrow 1^{-}} ( x^{2}+2x+ 3) =1^{2}+ 2(1)+ 3 = 6$

It must hold that $\lim_{x \rightarrow 1^{+}} f(x) =6$ as well; using the definition of f(x) for all positive values of :

$\lim_{x \rightarrow 1^{+}} f(x) = \lim_{x \rightarrow 1^{+}} (a+ b \ln x ) =a+ b \ln 1 = a + b \cdot 0 = a$ .

Therefore, a = 6 .

Now examine f'(x) . For to be differentiable, it must hold that

$\lim_{x \rightarrow 1^{-}} f'(x) = \lim_{x \rightarrow 1^{+}} f'(x)$

To find $\lim_{x \rightarrow 1^{-}} f'(x)$ , we can differentiate the expression for f(x) for all negative values of :

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

f'(x) = 2x + 2

Again, through straightforward substitution,

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

To find $\lim_{x \rightarrow 1^{+}} f'(x)$ , we can differentiate the expression for f(x) for all positive values of :

$f (x) = a+ b \ln x$

$f ' (x) = \frac{\mathrm{d} }{\mathrm{d} x}(a)+ \frac{\mathrm{d} }{\mathrm{d} x}(b \ln x)$

$f ' (x) = 0 +b \frac{\mathrm{d} }{\mathrm{d} x}( \ln x)$

$f ' (x) = b \cdot \frac{1}{x} = \frac{b}{x}$

Again, through substitution,

$\lim_{x \rightarrow 1^{+}} f'(x) = \lim_{x \rightarrow 1^{+}} \frac{b}{x} = b$

and b= 4 .

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Example Question #73 : Ap Calculus Ab

Find the derivative.

$12x^{12}-13x^2$

Possible Answers:

$144x^{11}+26x$

$144x^{11}-26x$

$144x^{11}+169x$

$x^{11}+x$

Correct answer:

$144x^{11}-26x$

Explanation:

Use the power rule to find the derivative.

$\frac{d}{dx}12x^{12}=144x^{11}$

$\frac{d}{dx}13x^2=26x$

Thus, the derivative is $144x^{11}-26x$

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Example Question #74 : Ap Calculus Ab

Find the derivative.

$\frac{x^2+3}{x^3}$

Possible Answers:

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

$\frac{-x^2-9}{x^3}$

$\frac{2x-3x^2-9x}{x^6}$

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

Correct answer:

$\frac{-x^2-9}{x^3}$

Explanation:

Use the quotient rule to find the derivative.

$\frac{x^3(2x)-(x^2+3)3x^2}{x^6}$

Simplify.

$\frac{2x^4-3x^4-9x^2}{x^6}=\frac{-x^2-9}{x^3}$

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Example Question #51 : Derivatives Of Functions

Find the first derivative of the function:

$h(x)=\csc(4x^3+2x)$

Possible Answers:

$-(12x^2+2)(\csc(4x^3+2x)\cot(4x^3+2x))$

$(12x^2+2)(\csc(4x^3+2x)\cot(4x^3+2x))$

$-(12x^2+2)(\csc(x)\cot(x))$

$-\csc(4x^3+2x)\cot(4x^3+2x)$

Correct answer:

$-(12x^2+2)(\csc(4x^3+2x)\cot(4x^3+2x))$

Explanation:

The derivative of the function is equal to

$h'(x)=-(12x^2+2)(\csc(4x^3+2x)\cot(4x^3+2x))$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \csc(x)=-\csc(x)\cot(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

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Example Question #77 : Derivatives

Given $f(x)=2x^{3}-7x^{2}+8x-3$ . Find f''(2) .

Possible Answers:

Correct answer:

Explanation:

First, find the first derivative.

You should get $f'(x)=6x^{2}-14x+8$ .

Next, differentiate again.

You should get f''(x)=12x-14 .

Finally, plug in x=2 to get f''(2)=10 .

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Example Question #78 : Derivatives

The velocity profile of a fluid is given by

$v(x)=16\cos(x)+x^2+16\sin(x)$

Determine the rate of change of the velocity of the fluid at any point.

Possible Answers:

$16(\cos(x)+\sin(x))+2x$

$16(\cos(x)-\sin(x))+2x$

$16(\cos(x)-\sin(x))+x^2$

2x+16

Correct answer:

$16(\cos(x)-\sin(x))+2x$

Explanation:

To find the rate of change of the velocity of the fluid at any point, we must take the derivative of the function:

$v'(x)=-16\sin(x)+2x+16\cos(x)=16(\cos(x)-\sin(x))+2x$

The derivative was found using the following rules:

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

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Example Question #79 : Derivatives

Find :

$f=\sqrt{\tan(4x)}$

Possible Answers:

$\frac{2\sec^2(4x)}{\sqrt{\tan(4x)}}$

$-\frac{2\sec^2(4x)}{\sqrt{\tan(4x)}}$

$\frac{4\sec^2(4x)}{\sqrt{\tan(4x)}}$

$\frac{1}{\sqrt{\tan(4x)}}$

Correct answer:

$\frac{2\sec^2(4x)}{\sqrt{\tan(4x)}}$

Explanation:

The derivative of the function is equal to

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

and was found using the following rules:

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

Note that the chain rule was used on the square root, the inner tangent function, and the function inside the secant.

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Example Question #80 : Derivatives

Find the derivative of the function:

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

Possible Answers:

$-2\cos(2x)\sin(2x)$

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

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

$-4\cos(4x)\sin(4x)$

Correct answer:

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

Explanation:

The derivative of the function is equal to

$f'=2\cos(2x)(-2\sin(2x))=-4\cos(2x)\sin(2x)$

and was found using the following rules:

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

Notice that the chain rule was used on the cosine squared function, the cosine function, and the function inside the sine function.

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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 #71 : Ap Calculus Ab

Example Question #72 : Ap Calculus Ab

Example Question #71 : Derivatives

Example Question #73 : Ap Calculus Ab

Example Question #74 : Ap Calculus Ab

Example Question #51 : Derivatives Of Functions

Example Question #77 : Derivatives

Example Question #78 : Derivatives

Example Question #79 : Derivatives

Example Question #80 : Derivatives

All AP Calculus AB Resources

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