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AP Calculus AB : Derivatives

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Example Questions

Example Question #81 : Derivatives Of Functions

Find the derivative of the function:

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

The derivative of the function is equal to

$f'=-2e^{2x}\sin(e^{2x})$

and was found using the following rules:

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

Note that the chain rule was used for both the inner function of cosine and the inner function of the exponential!

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Example Question #86 : Computation Of The Derivative

Find the function which gives the rate of change of f(x)

f(x)=16e^x+12x^6-48x+13

Possible Answers:

$f'(x)=16e^x+72x^{5}-48$

$f'(x)=16e^x+72x^{5}-48x+29$

$f'(x)=16e^x+18x^{5}-49x$

$f'(x)=16e^x+72x^{4}-48$

Correct answer:

$f'(x)=16e^x+72x^{5}-48$

Explanation:

Find the function which gives the rate of change of f(x)

f(x)=16e^x+12x^6-48x+13

Finding a function which models the rate of change of another function is the same thing as finding the derivative of that function.

To find our derivative, we need to recall two rules.

$\frac{dy}{dx}e^x=e^x$

And

$\frac{dy}{dx}x^n=nx^{n-1}$

Using these two rules, we can find the derivative of f(x).

Our first term can be derived using our first rule. The derivative of e to the x is just e to the x.

This means that our first term will remain 16e to x.

For our other three terms, we follow the second rule. We will decrease each term's exponent by 1, and then multiply the coefficient by the old exponent.

$f'(x)=16e^x+(6)12x^{6-1}-(1)48x^{1-1}$

Notice that the 13 will drop out. It is a constant term, and as such when we multiply it by it's original exponent (0) it wil be reduced to zero as well.

Clean up the above to get:

$f'(x)=16e^x+72x^{5}-48$

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Example Question #82 : Derivatives Of Functions

Find the first derivative of the function:

$f(x)=\ln(\sec(x^2))$

Possible Answers:

$-2x\tan(x^2)$

$2x\tan(x^2)$

$\tan(x^2)$

$2x\sec(x^2)\tan(x^2)$

Correct answer:

$2x\tan(x^2)$

Explanation:

The derivative of the function is equal to

$f'(x)=2x\tan(x^2)$

and was found using the following rules:

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

Note that the chain rule was used on the natural logarithm and the secant functions:

$f'(x)=\frac{2x\sec(x^2)\tan(x^2)}{\sec(x^2)}=2x\tan(x^2)$

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Example Question #83 : Derivatives Of Functions

Find the derivative of the function:

$f=\cos(x^2)^{\frac{3}{2}}$

Possible Answers:

$-3x\sin(x^2)\cos(x^2)$

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

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

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

Correct answer:

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

Explanation:

The derivative of the function is

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

and was found using the following rules:

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

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Example Question #84 : Derivatives Of Functions

Find the first derivative of the function:

$f=\tan^2(4x+2)$

Possible Answers:

$-8\sec^2(4x+2)\tan(4x+2)$

$8\sec^2(4x+2)\tan(4x+2)$

$4\sec^2(4x+2)\tan(4x+2)$

$8\sec(4x+2)\tan(4x+2)$

Correct answer:

$8\sec^2(4x+2)\tan(4x+2)$

Explanation:

The derivative of the function is equal to

$f'=8\sec^2(4x+2)\tan(4x+2)$

and was found using the following rules:

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

Note that the chain rule is used for the squared tangent and the function inside tangent.

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Example Question #85 : Derivatives Of Functions

Find the derivative of the function:

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

Possible Answers:

$e^x\sin(2x)(1+2\cos(2x))$

$e^x\sin(x)(1+\cos(x))$

$2e^x\sin(2x)$

$e^x\sin(2x)(1+\cos(2x))$

Correct answer:

$e^x\sin(2x)(1+2\cos(2x))$

Explanation:

The derivative of the function is equal to

$f'(x)=e^x\sin(2x)(1+2\cos(2x))$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^x=e^x$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \sin(x)=\cos(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

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Example Question #91 : Derivatives Of Functions

What is the second derivative of the following function?

f(x)=x^3-4x^2+5x+9

Possible Answers:

f''(x)=x-4

f''(x)=6x-8

f''(x)=3x+5

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

Correct answer:

f''(x)=6x-8

Explanation:

f(x)=x^3-4x^2+5x+9

f'(x)=3x^2-8x+5

f''(x)=6x-8

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Example Question #92 : Derivatives Of Functions

f(x)=6x^6-3x^4+x^3-12

What is f''(x) ?

Possible Answers:

f''(x)=36x^4-36x^2+9x

f''(x)=35x^5-12x^3+3x^2

f''(x)=160x^4-42x^2+9x

f''(x)=180x^4-36x^2+6x

f''(x)=180x^4-12x^3+6x

Correct answer:

f''(x)=180x^4-36x^2+6x

Explanation:

f(x)=6x^6-3x^4+x^3-12

f'(x)=6(6x^5)-4(3x^3)+3(x^2)-0=36x^5-12x^3+3x^2

f''(x)=5(36x^4)-3(12x^2)+2(3x)=180x^4-36x^2+6x

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Example Question #93 : Derivatives Of Functions

What function is equivalent to the tenth derivative of the following function?

$f(x)=514x^6\pm 59x^5\pm4x^4\pm36x^3\pm1$

Possible Answers:

$y=514(6!)x^5\pm59(5!)x^4\pm4(4!)x^3\pm36(3!)x^2$

The tenth derivative is undefined.

$y=185040x\pm25370$

$y=1228(x^{x-1})\pm514x^x$

y=0

Correct answer:

y=0

Explanation:

Even though the coefficients become massive as one progresses through successive derivatives of f(x), the sixth derivative yields a constant. The following seventh derivative will therefore equal zero, as will all following derivatives. We can confidently state — without necessarily deriving the equation ten times — that the tenth derivative will result in an equation equivalent to y=0.

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Example Question #93 : Derivatives Of Functions

h(x)=cos(2x)

Find $h''(\frac{\pi}{4} )$

Possible Answers:

$h''(\I\Pi/4 )=0$

None of the other answers.

$h''(\I\Pi/4 )=1$

$h''(\I\Pi/4 )=2$

$h''(\I\Pi/4 )=4$

Correct answer:

$h''(\I\Pi/4 )=0$

Explanation:

According to the chain rule:

if h(x)=f(g(x)) , then $h'(x)=f'(g(x))\cdot g'(x)$ .

So, $h'(x)=-sin(2x)\cdot2$ .

To find the second derivative, we will use the chain rule again:

$h''(x)=f''(g'(x))\cdot g''(x)$

f''(x)=-2cos(2x)

g''(x)=2

Therefore, $h''(x)=-2cos(2x)\cdot2$ $=-4cos(2x)\cdo$ .

Finally, we simply plug in $\frac{\pi}{4}$ into the second derivative.

$-4cos(2\cdot \frac{\pi}{4})=-4cos(\frac{\pi}{2})=-4(0)=0$

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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 #81 : Derivatives Of Functions

Example Question #86 : Computation Of The Derivative

Example Question #82 : Derivatives Of Functions

Example Question #83 : Derivatives Of Functions

Example Question #84 : Derivatives Of Functions

Example Question #85 : Derivatives Of Functions

Example Question #91 : Derivatives Of Functions

Example Question #92 : Derivatives Of Functions

Example Question #93 : Derivatives Of Functions

Example Question #93 : Derivatives Of Functions

All AP Calculus AB Resources

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