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AP Calculus AB : AP Calculus AB

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Example Questions

Example Question #1 : Chain Rule And Implicit Differentiation

Find the derivative of $f(x)=10^{2\sqrt{x}}$ .

Possible Answers:

$f'(x)=10^x^{{-{\frac{1}{2}}}}*ln10$

$f'(x)=10^{2\sqrt{x}}*2x^{-{\frac{1}{2}}}*ln10$

$f'(x)=10^{2\sqrt{x}}*ln10*x^{-{\frac{1}{2}}}$

$f'(x)=10^{2\sqrt{x}}*x^{-{\frac{1}{2}}}$

$f'(x)=ln10*x^{-{\frac{1}{2}}}$

Correct answer:

$f'(x)=10^{2\sqrt{x}}*ln10*x^{-{\frac{1}{2}}}$

Explanation:

To find the derivative, we can first rewrite the function to make it easier to take the chain rule. Rewrite $10^{2\sqrt{x}}$ as $10^{2x^{\frac{1}{2}}}$ . Now, like in any exponential function, the first factor of the derivative is the original exponential function. So, the first factor of f'(x) will be $10^{2\sqrt{x}}$ . Next, by the chain rule for derivatives, we must take the derivative of the exponent, which is why we rewrote the exponent in a way that is easier to take the derivative of. So, the derivative of the exponent is $x^{-{\frac{1}{2}}}$ , because the 1/2 and the 2 cancel when we bring the power down front, and the exponent of 1/2 minus 1 becomes negative 1/2. The last factor of the derivative is ln10 because in every derivative of an exponential function where the base is a number, we must multiply by the natural log of that base. So, once you multiply all these factors together, the final answer is $f'(x)=10^{2\sqrt{x}}*ln10*x^{-{\frac{1}{2}}}$

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Example Question #2 : Chain Rule And Implicit Differentiation

If x^3+y^3=6xy , find the derivative through implicit differentiation.

Possible Answers:

$y'=\frac{2y+2x}{x-y}$

$y'=\frac{2y-x^2}{y^2-2x}$

$y'=\frac{y-x^2}{y^2-x}$

$y'=\frac{3y-6x^2}{6y^2-3x}$

$y'=\frac{2y+x^2}{y^2+2x}$

Correct answer:

$y'=\frac{2y-x^2}{y^2-2x}$

Explanation:

To find the derivative through implicit differentiation, we have to take the derivative of every term with respect to x. Don't forget that each time you take the derivative of a term containing y, you must multiply its derivative by y'. So, when we take the derivative of each term, we get 3x^2+3y^2(y')=6y+6x(y'). The next step is to solve for y', so we put all terms containing y' on the left side of the equation: 3y^2(y')-6x(y')=6y-3x^2 . Next, factor out the y' from both terms on the left side of the equation so that we can solve for it: y'(3y^2-6x)=6y-3x^2. To get y' alone, divide both sides by to get $y'=\frac{6y-3x^2}{3y^2-6x}$ . To simplify even further, we can factor a 2 out of the numerator and denominator and cancel them. So, the final answer is $y'=\frac{2y-x^2}{y^2-2x}$ .

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Example Question #183 : Derivatives

If sin(x+y)=y^2(cosx) , find .

Possible Answers:

$y'=\frac{cos(x+y)-y^2(sinx)}{2ycosx-cos(x+y)}$

$y'=\frac{-2ysinx-cosx}{cos(x+y)-y^2cosx}$

$y'=tanxy-\frac{cos(x+y)}{2ycosx}$

$y'=\frac{-y^2(sinx)-cos(x+y)}{cos(x+y)-2ycosx}$

$y'=\frac{-2ycosx(sinx)}{cos(x+y)}$

Correct answer:

$y'=\frac{-y^2(sinx)-cos(x+y)}{cos(x+y)-2ycosx}$

Explanation:

To find the derivative through implicit differentiation, we have to take the derivative of every term with respect to x. Don't forget that each time you take the derivative of a term containing y, you must multiply its derivative by y'. So, when we take the derivative of each term, we get cos(x+y)(1+y')=y^2(-sinx)+cosx(2y*y'). The next step is to solve for y', so we put all terms containing y' on the left side of the equation and factor out a common y': y'(cos(x+y)-2ycosx)=-y^2sinx-cos(x+y) . To get y' alone, divide both sides by to get $y'=\frac{-y^2sinx-cos(x+y)}{cos(x+y)-2ycosx}$ .

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Example Question #1 : Chain Rule And Implicit Differentiation

Find the derivative of the function of the circle x^2+y^2=25.

Possible Answers:

$\frac{-x}{y}$

$\frac{x}{y}$

$\frac{-y}{x}$

Correct answer:

$\frac{-x}{y}$

Explanation:

To find the derivative through implicit differentiation, we have to take the derivative of every term with respect to x. Don't forget that each time you take the derivative of a term containing y, you must multiply its derivative by y'. So, when we take the derivative of each term, we get 2x+2y(y')=0. The next step is to solve for y', so we put all terms containing y' on the left side of the equation: 2y(y')=-2x. . To get y' alone, divide both sides by to get $y'=\frac{-2x}{2y}$ . To simplify even further, we can factor a 2 out of the numerator and denominator and cancel them. So, the final answer is $\frac{-x}{y}$ .

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Example Question #184 : Derivatives

Find the derivative of the function 4x^3-3y=8 using implicit differentiation.

Possible Answers:

$y'=\frac{8-4x^3}{-3}$

y'=4x^2

$y'=\frac{1}{4}x^2$

y'=-4x^2

cannot be solved

Correct answer:

y'=4x^2

Explanation:

To find the derivative through implicit differentiation, we have to take the derivative of every term with respect to x. Don't forget that each time you take the derivative of a term containing y, you must multiply its derivative by y'. So, when we take the derivative of each term, we get 12x^2-3y'=0 . The next step is to solve for y', so we put all terms containing y' on the left side of the equation: -3y'=-12x^2 . To get y' alone, divide both sides by -3 to get $y'=\frac{-12x^2}{-3}$ . To simplify even further, we can factor a -2 out of the numerator and denominator and cancel them. So, the final answer is y'=4x^2 .

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Example Question #2 : Chain Rule And Implicit Differentiation

Find the derivative of the function x^2y+2y^3=3x+2y .

Possible Answers:

$y'=\frac{x^2+6y^2-2}{3-2xy}$

$y'=\frac{3-2xy}{x^2+6y^2-2}$

$y'=\frac{5}{x^2+6y^2}$

$y'=\frac{2-3xy}{x^2+2y^2-6}$

$y'=\frac{3+2xy}{x^2+2-6y^2}$

Correct answer:

$y'=\frac{3-2xy}{x^2+6y^2-2}$

Explanation:

To find the derivative through implicit differentiation, we have to take the derivative of every term with respect to x. Don't forget that each time you take the derivative of a term containing y, you must multiply its derivative by y'. So, when we take the derivative of each term, we get 2x(y)+(x^2)y'+(6y^2)y'=3+2y' . The next step is to solve for y', so we put all terms containing y' on the left side of the equation: x^2y'+6y^2(y')-2y'=3-2xy . Next, factor out the y' from both terms on the left side of the equation so that we can solve for it: y'(x^2+6y^2-2)=3-2xy . To get y' alone, divide both sides by to get a final answer of $y'=\frac{3-2xy}{x^2+6y^2-2}$ .

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Example Question #185 : Derivatives

Find the derivative of the function lny=lnx^6 .

Possible Answers:

$y'=\frac{6y}{x}$

$y'=\frac{x}{6y}$

undefined

$y'=\frac{6x}{y}$

$y'=\frac{y}{6x}$

Correct answer:

$y'=\frac{6y}{x}$

Explanation:

Before we take the derivative of the logarithmic function, we can make it easier for ourselves by simplifying the equation to lny=6lnx . We can bring the exponent of 6 down in front of the natural log of x due to properties of logarithms. Next, take the derivative of each term in terms of x. Don't forget to multiply by y' each time you take the derivative of a term containing y! When we do this, we should get $(\frac{1}{y})y'=6(\frac{1}{x})$ because the derivative of lnx is 1/x. Next, solve for y' by multiplying both sides by y to get the final answer of $y'=\frac{6y}{x}$ .

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Example Question #101 : Computation Of The Derivative

$f(x)=sin(x^{2}})$ . Find f'(x) .

Possible Answers:

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

f'(x)=-2xcos(x^2)

f'(x)=2xcos(x)

f'(x)=2xsin(x^2)

$f'(x)=2xcos(x^{2}})$

Correct answer:

$f'(x)=2xcos(x^{2}})$

Explanation:

To take the derivative, you must first take the derivative of the outside function, which is sine. However, the x^2 , or the angle of the function, remains the same until we take its derivative later. The derivative of sinx is cosx, so you the first part of f'(x) will be cos(x^2) . Next, take the derivative of the inside function, x^2 . Its derivative is , so by the chain rule, we multiply the derivatives of the inside and outside functions together to get f'(x)=2xcos(x^2) .

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Example Question #102 : Computation Of The Derivative

f(x)=(4x^3-9x^2+4x-8)^2 . Using the chain rule for derivatives, find f'(x) .

Possible Answers:

f'(x)=12x^2-18x+4

f'(x)=2(4x^3-9x^2+4x-8)

f'(x)=8x^5-18x^4+8x^2-16

f'(x)=(12x^2-18x+4)^2

f'(x)=2(4x^3-9x^2+4x-8)^2*(12x^2-18x+4)

Correct answer:

f'(x)=2(4x^3-9x^2+4x-8)^2*(12x^2-18x+4)

Explanation:

By the chain rule, we must first take the derivative of the outside function by bringing the power down front and reducing the power by one. When we do this, we do not change the function that is in the parentheses, or the inside function. That means that the first part of f'(x) will be 2(4x^3-9x^2+4x-8) . Next, we must take the derivative of the inside function. Its derivative is (12x^2-18x+4) . The chain rule says we must multiply the derivative of the outside function by the derivative of the inside function, so the final answer is f'(x)=2(4x^3-9x^2+4x-8)*(12x^2-18x+4) .

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Example Question #13 : Chain Rule And Implicit Differentiation

Use implicit differentiation to find $\frac{dy}{dx}$ is terms of and for,

$xy^2=y\sin(x)+1$

Possible Answers:

$\frac{dy}{dx}=\frac{\sin(x)-3}{2x +\cos(x)}$

$\frac{dy}{dx}=\frac{y(\cos x-y)}{2xy -\sin x}$

$\frac{dy}{dx}=\frac{2y\sin(x)-3y^2}{2y -\cos(x)}$

$\frac{dy}{dx}=\frac{2\sin(x)+3y^2}{2y -x\sin(x)}$

$\frac{dy}{dx}=\frac{\sin(x)-1}{2xy +\cos(x)}$

Correct answer:

$\frac{dy}{dx}=\frac{y(\cos x-y)}{2xy -\sin x}$

Explanation:

$xy^2=y\sin(x)+1$

To differentiate the equation above, start by applying the derivative operation to both sides,

$\frac{d}{dx}(xy^2)=\frac{d}{dx}[y\sin(x)]+\underset{=0}{ \underbrace{\frac{d}{dx}(1)}}$

Both sides will require the product rule to differentiate,

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

$y^2\underset{=1}{ \underbrace{\frac{d}{dx}(x)}} +x\left(2y\frac{dy}{dx} \right ) =\sin(x)\frac{dy}{dx}+y\cos(x)$

________________________________________________________________

Common Mistake

A common mistake in the previous step would be to conclude that $\frac{d}{dx}(y^2)=2y$ instead of $2y\frac{dy}{dx}$ . The former is not correct; if we were looking for the derivative with respect to , then $\frac{d}{dy}(y^2)$ would in fact be . But we are not differentiating with respect to , we're looking for the derivative with respect to .

We are assuming that is a function of , so we must apply the chain rule by differentiating with respect to and multiplying by the derivative of with respect to to obtain $\frac{d}{dx}(y^2)=2y\frac{dy}{dx}$ .

________________________________________________________________

$y^2+2xy\frac{dy}{dx} =\sin(x)\frac{dy}{dx}+y\cos(x)$

Collect terms with a derivative onto one side of the equation, factor out the derivative, and divide out to solve for the derivative $\frac{dy}{dx}$ ,

$2xy\frac{dy}{dx} -\sin(x)\frac{dy}{dx}=y\cos(x)-y^2$

$\frac{dy}{dx}[2xy -\sin(x)]=y\cos(x)-y^2$

$\frac{dy}{dx}=\frac{y\cos(x)-y^2}{2xy -\sin(x)}=\frac{y(\cos x-y)}{2xy -\sin (x)}$

Therefore,

$\frac{dy}{dx}=\frac{y(\cos x-y)}{2xy -\sin x}$

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AP Calculus AB : AP Calculus AB

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #1 : Chain Rule And Implicit Differentiation

Example Question #2 : Chain Rule And Implicit Differentiation

Example Question #183 : Derivatives

Example Question #1 : Chain Rule And Implicit Differentiation

Example Question #184 : Derivatives

Example Question #2 : Chain Rule And Implicit Differentiation

Example Question #185 : Derivatives

Example Question #101 : Computation Of The Derivative

Example Question #102 : Computation Of The Derivative

Example Question #13 : Chain Rule And Implicit Differentiation

All AP Calculus AB Resources

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