AP Calculus AB : AP Calculus AB

Study concepts, example questions & explanations for AP Calculus AB

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Example Questions

Example Question #741 : Ap Calculus Ab

Find the derivative of the function

\displaystyle f=12x^2+\sin(x)-100

Possible Answers:

\displaystyle 24x^2-\cos(x)

\displaystyle 24x-\cos(x)

\displaystyle 12x+\cos^2(x)

\displaystyle 24x+\cos(x)

Correct answer:

\displaystyle 24x+\cos(x)

Explanation:

To find the derivative of the sum, we take the derivative of each term independently, then add them all up. Further, we use the rule

\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(x^n)=nx^{n-1}

\displaystyle f'=\frac{\mathrm{d} }{\mathrm{d} x}(12x^2)+\frac{\mathrm{d} }{\mathrm{d} x}(\sin(x))-\frac{\mathrm{d} }{\mathrm{d} x}(100)

\displaystyle f'=24x+\cos(x)

Example Question #741 : Ap Calculus Ab

Find the derivative of the following function:

\displaystyle f=2x^2+5x-1

Possible Answers:

\displaystyle f'=4x+4

\displaystyle f'=9x

\displaystyle f'=4x^3+5x^2

\displaystyle f'=4x+5

Correct answer:

\displaystyle f'=4x+5

Explanation:

To find the derivative of the function, we take the derivative of each element in the function independently, then add them up.

Using \displaystyle \frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}, we solve

\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(2x^2)+\frac{\mathrm{d} }{\mathrm{d} x}(5x)+\frac{\mathrm{d} }{\mathrm{d} x}(-1)=4x+5

Example Question #55 : Derivative Rules For Sums, Products, And Quotients Of Functions

Find the derivative of the following function:

\displaystyle f=2x^5\sin{x}

Possible Answers:

\displaystyle f'=10x^4\sin{x}+2x^5\cos{x}

\displaystyle f'=10x^4\sin{x}+2x^5\sin{x}

\displaystyle f'=10x^4\sin{x}-2x^5\cos{x}

\displaystyle f'=10x^5\sin{x}+x^5\cos{x}

Correct answer:

\displaystyle f'=10x^4\sin{x}+2x^5\cos{x}

Explanation:

To find the derivative of the function, we use the product rule, which is defined as

\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(f*g)=(f'*g)+(f*g'), where f and g are both functions.

\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(2x^5\sin{x})=\frac{\mathrm{d} }{\mathrm{d} x}(2x^5)*(\sin{x})+\frac{\mathrm{d} }{\mathrm{d} x}(\sin{x})*(2x^5)=10x^4\sin{x}+2x^5\cos{x}

Example Question #261 : Computation Of The Derivative

Find the derivative of the following function using the quotient rule:

\displaystyle f=\frac{e^x}{\tan{x}}

Possible Answers:

\displaystyle f'=\frac{xe^x\tan{x}-e^x\sec^2{x}}{\tan^2{x}}

\displaystyle f'=\frac{e^x\tan{x}-e^x\sec^2{x}}{\tan^2{x}}

\displaystyle f'=\frac{e^x\tan{x}-e^x\sec^2{x}}{\tan{x}}

\displaystyle f'=\frac{e^x\tan{x}+e^x\sec^2{x}}{\tan^2{x}}

Correct answer:

\displaystyle f'=\frac{e^x\tan{x}-e^x\sec^2{x}}{\tan^2{x}}

Explanation:

To find the derivative of the function using the quotient rule, we apply the following definition:

\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(\frac{f}{g})=\frac{gf'-fg'}{g^2}

\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(\frac{e^x}{\tan{x}})=\frac{\tan{x}*\frac{\mathrm{d} }{\mathrm{d} x}(e^x)-e^x*\frac{\mathrm{d} }{\mathrm{d} x}(\tan{x})}{\tan^2{x}}=\frac{e^x\tan{x}-e^x\sec^2{x}}{\tan^2{x}}

 

 

 

Example Question #742 : Ap Calculus Ab

Find the derivative of the following function:

\displaystyle f=10x^3+5x^2-1

Possible Answers:

\displaystyle f'=3x^2+x

\displaystyle f'=30x^3+10x^2

\displaystyle f'=30x^2+10x

\displaystyle f'=30x^2+10x-1

Correct answer:

\displaystyle f'=30x^2+10x

Explanation:

To find the derivative of the function, we take the derivative of each element in the function independently, then add them up.

Using \displaystyle \frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}, we solve

\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(10x^3)+\frac{\mathrm{d} }{\mathrm{d} x}(5x^2)+\frac{\mathrm{d} }{\mathrm{d} x}(-1)=30x^2+10x

Example Question #54 : Derivative Rules For Sums, Products, And Quotients Of Functions

Determine the second derivative of \displaystyle f(x)=ln(x^2-1)

Possible Answers:

\displaystyle f''(x)=\frac{-2x^2-2}{x^4-2x^2+1}

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

\displaystyle f''(x)=\frac{2x^2-1}{x^4-2x^2}

\displaystyle f''(x)=\frac{x}{2x -1}

\displaystyle f''(x)=\frac{2x^2-2}{x^3-x^2+1}

Correct answer:

\displaystyle f''(x)=\frac{-2x^2-2}{x^4-2x^2+1}

Explanation:

Finding our second derivative requires two steps, we first must find the derivative then find the corresponding rate of change for that new equation.

\displaystyle f(x)=ln(x^2-1)

Here, the chain rule is used since our function is of the form \displaystyle [f(g(x))]dx=f'(x)g'(f(x))

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

We now must use the quotient rule since our function is a rational function. We use the rule \displaystyle [\frac{f(x)}{g(x)}]dx=\frac{f'(x)g(x)-g'(x)f(x))}{(g(x))^2}

Therefore,

\displaystyle f''(x)=\frac{-2x^2-2}{x^4-2x^2+1}

 

Example Question #741 : Ap Calculus Ab

What is the derivative of \displaystyle x^3+2x+5?

Possible Answers:

\displaystyle 3x^2+2

\displaystyle 3x^2+2x

\displaystyle 25

\displaystyle 5

\displaystyle \frac{1}{3}x^2+\frac{1}{2}x+\frac{1}{5}

Correct answer:

\displaystyle 3x^2+2

Explanation:

To solve this problem, we can use the power rule. That means we lower the exponent of the variable by one and multiply the variable by that original exponent.

We're going to treat \displaystyle 5 as \displaystyle 5x^0, as anything to the zero power is one.

\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x+5)=(3*x^{3-1})+(1*2x^{1-1})+(0*5x^{0-1})

Notice that \displaystyle (0*5x^{0-1})=0, as anything times zero is zero.

\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x+5)=(3*x^{3-1})+(1*2x^{1-1})

\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x+5)=(3*x^{2})+(2x^{0})

\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x+5)=3x^2+2

Example Question #743 : Ap Calculus Ab

Find the slope of the f(x)=\sqrt{x^2 +2x-3}\displaystyle f(x)=\sqrt{x^2 +2x-3} at \displaystyle x=2.

Possible Answers:

\frac{\sqrt{5}}{5}\displaystyle \frac{\sqrt{5}}{5}

\frac{3\sqrt{5}}{5}\displaystyle \frac{3\sqrt{5}}{5}

\frac{6\sqrt{5}}{5}\displaystyle \frac{6\sqrt{5}}{5}

6\displaystyle 6

5\displaystyle 5

Correct answer:

\frac{3\sqrt{5}}{5}\displaystyle \frac{3\sqrt{5}}{5}

Explanation:

First we need to find the derivative of the function. f'(x)=\frac{1+x}{\sqrt{-3+2x+x^2}}\displaystyle f'(x)=\frac{1+x}{\sqrt{-3+2x+x^2}}

Now, we can plug in \displaystyle x=2 to the derivative function.

f'(2)=\frac{3\sqrt{5}}{5}\displaystyle f'(2)=\frac{3\sqrt{5}}{5}

Example Question #63 : Derivative Rules For Sums, Products, And Quotients Of Functions

Find the derivative of the following function:

\displaystyle f=\frac{x^4\sin{x}}{x^2+3}

Possible Answers:

\displaystyle f'=\frac{(x^2+3)(4x^3\sin(x)+x^4\cos(x)-2x^5\sin(x))}{(x^2+3)^2}

\displaystyle f'=\frac{(x^2+3)(4x^3\sin(x)+x^4\cos(x)-2x^5\cos(x))}{(x^2+3)^2}

\displaystyle f'=\frac{(x^2+3)(4x^3\sin(x)+x^4\cos(x)-2x^5\sin(x))}{(x^2-3)^2}

\displaystyle f'=\frac{(x^2-3)(4x^3\sin(x)+x^4\cos(x)+2x^5\sin(x))}{(x^2+3)^2}

Correct answer:

\displaystyle f'=\frac{(x^2+3)(4x^3\sin(x)+x^4\cos(x)-2x^5\sin(x))}{(x^2+3)^2}

Explanation:

To find the derivative of the function, we use a combination of the quotient rule and product rule

\displaystyle f=\frac{x^4\sin{x}}{x^2+3}

\displaystyle f'=\frac{(x^2+3)(x^4\sin{x})'-x^4\sin{x}(x^2+3)'}{(x^2+3)^2}

\displaystyle f'=\frac{(x^2+3)(4x^3\sin{x}+x^4\cos{x})-x^4\sin{x}(2x)}{(x^2+3)^2}

\displaystyle f'=\frac{(x^2+3)(4x^3\sin(x)+x^4\cos(x)-2x^5\sin(x))}{(x^2+3)^2}

Example Question #63 : Derivative Rules For Sums, Products, And Quotients Of Functions

Find the derivative of the function 

\displaystyle f=x^3+5x^2+8x+20

Possible Answers:

\displaystyle f'=\frac{3x^2+10x^2+8}{x}

\displaystyle f'=3x^3+10x^2+8x

\displaystyle f'=3x^2+10x+8

\displaystyle f'=3x^2-10x+8

Correct answer:

\displaystyle f'=3x^2+10x+8

Explanation:

To find the derivative of the function, we use the rule

\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1} and apply it to each term in the function

\displaystyle f'=\frac{\mathrm{d} }{\mathrm{d} x}(x^3)+\frac{\mathrm{d} }{\mathrm{d} x}(5x^2)+\frac{\mathrm{d} }{\mathrm{d} x}(8x)+\frac{\mathrm{d} }{\mathrm{d} x}(20)

\displaystyle f'=3x^2+10x+8

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