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AP Calculus AB : Computation of the Derivative

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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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Example Question #81 : Derivatives Of Functions

Find the derivative of the function:

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

Possible Answers:

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

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

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

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

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 #82 : Derivatives Of Functions

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

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

Possible Answers:

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

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

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

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

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 #83 : Derivatives Of Functions

Find the first derivative of the function:

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

Possible Answers:

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

$2x\tan(x^2)$

$\tan(x^2)$

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

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 #81 : Computation Of The Derivative

Find the derivative of the function:

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

Possible Answers:

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

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

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

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

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 #81 : Derivatives Of Functions

Find the first derivative of the function:

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

Possible Answers:

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

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

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

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

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 #86 : Derivatives Of Functions

Find the derivative of the function:

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

Possible Answers:

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

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

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

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

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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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 #97 : Ap Calculus Ab

Example Question #98 : Ap Calculus Ab

Example Question #99 : Ap Calculus Ab

Example Question #100 : Ap Calculus Ab

Example Question #81 : Derivatives Of Functions

Example Question #82 : Derivatives Of Functions

Example Question #83 : Derivatives Of Functions

Example Question #81 : Computation Of The Derivative

Example Question #81 : Derivatives Of Functions

Example Question #86 : Derivatives Of Functions

All AP Calculus AB Resources

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