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AP Calculus AB : Derivatives

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Example Questions

Example Question #231 : Derivatives

Find $\frac{\mathrm{d}y }{\mathrm{d} x}$ of the following equation:

x^2y+y^2=x

Possible Answers:

$\frac{1+2x}{2y-x^2}$

2xy+2y-1

x^2+2y

$\frac{1+2x}{2y+x^2}$

$\frac{1-2x}{2y+x^2}$

Correct answer:

$\frac{1-2x}{2y+x^2}$

Explanation:

To find $\frac{\mathrm{d}y }{\mathrm{d} x}$ we must use implicit differentiation, which is an application of the chain rule.

Taking $\frac{\mathrm{d} }{\mathrm{d} x}$ of both sides of the equation, we get

$2x+x^2\frac{\mathrm{d} y}{\mathrm{d} x}+2y\frac{\mathrm{d} y}{\mathrm{d} x}=1$

using the following rules:

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

Note that for every derivative of a function with y, the additional term $\frac{\mathrm{d} y}{\mathrm{d} x}$ appears; this is because of the chain rule, where y = g(x) , so to speak, for the function it appears in.

Using algebra to rearrange, we get

$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{1-2x}{2y+x^2}$

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Example Question #141 : Computation Of The Derivative

Find the derivative:

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

The derivative of the function is equal to

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

and was found using the following rules:

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

Before simplification, the derivative we get is

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

Note that the square root, the exponential, and the tangent function all utilize chain rule when taking their derivatives.

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Example Question #446 : Ap Calculus Ab

Find the first derivative of the following function:

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

Possible Answers:

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

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

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

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

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 first rule - the chain rule - was used three times for the function: the cosine, contained the exponential which itself was raised to a function. (Note that sometimes the exponential rule is written as $\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u \frac{\mathrm{d} u}{\mathrm{d} x}$ , which itself is the chain rule.)

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Example Question #51 : Chain Rule And Implicit Differentiation

A reaction is modeled by the following equation:

$\kappa (T)=Ae^{\frac{-R}{T}}$

where A, R are constants.

What is $\kappa '$ ?

Possible Answers:

$\frac{AR}{T}e^{-\frac{R}{T}}$

$-\frac{AR}{T^2}e^{-\frac{R}{T}}$

$\frac{AR}{T^2}e^{-\frac{R}{T}}$

Correct answer:

$\frac{AR}{T^2}e^{-\frac{R}{T}}$

Explanation:

The derivative of the function is equal to

$\kappa'(T)=\frac{AR}{T^2}e^{-\frac{R}{T}}$

and was found using the following rules:

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

The chain rule was used for the function contained in the exponential function (or, as written as a rule, $\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u \frac{\mathrm{d} u}{\mathrm{d} x}$ ).

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Example Question #141 : Computation Of The Derivative

For the equation, $y^2 + 2x^2y - \frac{5x}{y} = 2$ , find $\frac{\mathrm{d} y}{\mathrm{d} x}$ .

Possible Answers:

$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{5y^3 - 4xy^3}{2y^4 + 2x^2y^2 + 5x}$

$\frac{\mathrm{d} y}{\mathrm{d} x} =0$

$\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{4xy^3 - 5y}{2y^4 + 2x^2y^2 + 5x}$

$\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{ - 4xy^3 + 5y}{2y^3 + 2x^2y^2 +5x}$

Correct answer:

$\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{ - 4xy^3 + 5y}{2y^3 + 2x^2y^2 +5x}$

Explanation:

The equation given is not written and y = f(x) . Instead, it is written with 's and 's on the same side of the equation. This suggests we should try implicit differentiation, which means find the derivative of both sides with respect to .

"With respect to " means that we treat every other variable as a function of . So y = f(x) , and the derivative of is a chain rule. This will be emphasized later in the explanation.

First we must differentiate both sides with respect to .

$\frac{\mathrm{d} }{\mathrm{d} x}[y^2 + 2x^2y - \frac{5x}{y}] = \frac{\mathrm{d} }{\mathrm{d} x}[2]$

We have multiple terms on the left hand side, so we will differentiate each term individually.

$\frac{\mathrm{d} }{\mathrm{d} x}[y^2] + \frac{\mathrm{d} }{\mathrm{d} x}[2x^2y] + \frac{\mathrm{d} }{\mathrm{d} x}[\frac{-5x}{y}] = \frac{\mathrm{d} }{\mathrm{d} x}[2]$

For the first term, $\frac{\mathrm{d} }{\mathrm{d} x}[y^2]$ , we will use the power rule and also use the chain rule, since we must assume that y = f(x) .

$\frac{\mathrm{d} }{\mathrm{d} x}[y^2]$

${\color{Blue} 2[y]^{2-1}}{\color{DarkRed} (\frac{\mathrm{d} y}{\mathrm{d} x})}$

${\color{Blue} 2y}{\color{DarkRed} \frac{\mathrm{d} y}{\mathrm{d} x}}$

The blue part is the power rule. The red is from the chain rule. The red part is the derivative of y with respect to x, which is currently unknown. Remember that y is some unknown function of x, whose derivative is also unknown. We can only write $\frac{\mathrm{d} y}{\mathrm{d} x}$ for the derivative of y.

Now we find the second term's derivative,

$\frac{\mathrm{d} }{\mathrm{d} x}[{\color{Blue} 2x^2}{\color{DarkRed} y}]$

This is a product rule. To help with the product rule, the two pieces are color coded. Remember that the power rule is $\frac{\mathrm{d} }{\mathrm{d} x}[{\color{Blue} f(x)}\cdot{\color{DarkRed} g(x)}] ={\color{Blue} f(x)}\cdot {\color{DarkRed} g'(x)} + {\color{Blue} f'(x)} \cdot {\color{DarkRed} g(x)}$

Applying this, we get

${\color{Blue} (2x^2)}{\color{DarkRed} (\frac{\mathrm{d} y}{\mathrm{d} x})} + {\color{Blue} (4x)}{\color{DarkRed} (y)}$

Simplifying gives us

$2x^2 \cdot \frac{\mathrm{d} y}{\mathrm{d} x} + 4xy$

For the third term, $\frac{-5x}{y}$ , we will algebraically rewrite it as $-5xy^{-1}$ , so we can apply the product rule instead of the quotient rule. This is just a personal preference, The quotient rule would work as well.

$\frac{\mathrm{d} }{\mathrm{d} x}[{\color{Blue} -5x}{\color{DarkRed} y^{-1}}]$

${\color{Blue} (-5x)}{\color{DarkRed} (-1y^{-1-1})(\frac{\mathrm{d} y}{\mathrm{d} x})} + {\color{Blue} (-5)}{\color{DarkRed} (y^{-1})}$

remember that the derivative of ${\color{DarkRed} y^{-1}}$ with respect to requires the chain rule, resulting in the ${\color{DarkRed} (\frac{\mathrm{d} y}{\mathrm{d} x})}$ .Simplifying gives us

$5xy^{-2}\frac{\mathrm{d} y}{\mathrm{d} x} - 5y^{-1}$

The right hand side of the equation is a constant, so its derivative is zero.

Assembling all the parts back together, we have

$2y\frac{\mathrm{d} y}{\mathrm{d} x} + 2x^2\frac{\mathrm{d} y}{\mathrm{d} x} + 4xy + 5xy^{-2}\frac{\mathrm{d} y}{\mathrm{d} x} - 5y^{-1 }=0$

Now that we have differentiated the equation, we need to algebraically solve for $\frac{\mathrm{d} y}{\mathrm{d} x}$ .

First, we should move all terms with a $\frac{\mathrm{d} y}{\mathrm{d} x}$ to one side, which in this case is already done. Then we should move all terms without a $\frac{\mathrm{d} y}{\mathrm{d} x}$ to the opposite side of the equation. Doing so, we get

$2y\frac{\mathrm{d} y}{\mathrm{d} x} + 2x^2\frac{\mathrm{d} y}{\mathrm{d} x} + 5xy^{-2}\frac{\mathrm{d} y}{\mathrm{d} x} = -4xy+ 5y^{-1}$

Then we will will factor out the common factor of $\frac{\mathrm{d} y}{\mathrm{d} x}$ . This results in

$\frac{\mathrm{d} y}{\mathrm{d} x}[2y + 2x^2 + 5xy^{-2}] = -4xy + 5y^{-1}$

Then, to isolate $\frac{\mathrm{d} y}{\mathrm{d} x}$ , we divide both sides by $[2y + 2x^2 + 5xy^{-2}]$ .

$\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{-4xy + 5y^{-1}}{2y + 2x^2 + 5xy^{-2}}$

Now we need to simplify and get rid of the negative exponents. To do this, we can simply multiply the numerator and denominator by y^2 . This will result in

$\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{-4xy + 5y^{-1}}{2y + 2x^2 + 5xy^{-2}} \cdot \frac{y^2}{y^2}$

$\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{-4xy^3 + 5y}{2y^3 + 2x^2y^2 + 5x}$

Which is the correct answer.

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Example Question #453 : Ap Calculus Ab

Given the function $\small \small \small z(x)=\cos x^{12}$ , find its derivative.

Possible Answers:

$\small \small \small \small z'(x)=12x^{11}\sin x^{12}$

$\small \small \small \small z'(x)=-12x^{11}\cos x^{12}$

$\small \small \small z'(x)=-12x^{11}\sin x^{12}$

$\small \small \small \small z'(x)=-\sin 12x^{11}$

Correct answer:

$\small \small \small z'(x)=-12x^{11}\sin x^{12}$

Explanation:

Given the function $\small \small \small z(x)=\cos x^{12}$ , we can find its derivative using the chain rule, which states that

$\small [f(g(x))]'=f'(g(x))g'(x)$

where $\small \small f(x)=\cos x$ and $\small \small g(x)=x^{12}$ for $\small z(x)$ . We have $\small f'(x)=-\sin x$ and $\small g'(x)=12x^{11}$ , which gives us

$\small \small z'(x)=-\sin x^{12}\cdot 12x^{11}=-12x^{11}\sin x^{12}$

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Example Question #454 : Ap Calculus Ab

Given the function $\small \small \small \small z(x)=e^{x^{-1}}$ , find its derivative.

Possible Answers:

$\small \small \small \small \small z'(x)=\frac{e^{x^{-1}}}{x}$

$\small \small \small \small z'(x)=-\frac{e^{x^{-1}}}{x^2}$

$\small \small \small \small \small z'(x)=-\frac{e^{x^{-1}}}{x}$

$\small \small \small \small \small z'(x)=e^{-x^{-2}}$

Correct answer:

$\small \small \small \small z'(x)=-\frac{e^{x^{-1}}}{x^2}$

Explanation:

Given the function $\small \small \small \small z(x)=e^{x^{-1}}$ , we can find its derivative using the chain rule, which states that

$\small [f(g(x))]'=f'(g(x))g'(x)$

where $\small \small \small f(x)=e^{x}$ and $\small \small \small g(x)=x^{-1}$ for $\small z(x)$ . We have $\small \small f'(x)=e^x$ and $\small \small g'(x)=-x^{-2}$ , which gives us

$\small \small \small z'(x)=e^{x^{-1}}\cdot (-x^{-2})=-\frac{e^{x^{-1}}}{x^2}$

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Example Question #455 : Ap Calculus Ab

Given the function $\small \small \small z(x)=e^{-x+x^3}$ , find its derivative.

Possible Answers:

$\small \small \small \small \small z'(x)=e^{-x+x^3}\cdot (-x+x^{3})$

$\small \small \small \small z'(x)=e^{-x+x^3}\cdot (-1+x^{3})$

$\small \small \small \small \small \small \small z'(x)=e^{-x+x^3}\cdot (-1+3x^{2})$

$\small \small \small \small \small z'(x)=e^{-1+3x^2}\cdot (-1+x^{3})$

Correct answer:

$\small \small \small \small \small \small \small z'(x)=e^{-x+x^3}\cdot (-1+3x^{2})$

Explanation:

Given the function $\small \small \small z(x)=e^{-x+x^3}$ , we can find its derivative using the chain rule, which states that

$\small [f(g(x))]'=f'(g(x))g'(x)$

where $\small \small \small f(x)=e^{x}$ and $\small \small \small \small g(x)=-x+x^3$ for $\small z(x)$ . We have $\small \small f'(x)=e^x$ and $\small \small \small g'(x)=-1+3x^2$ , which gives us

$\small \small \small \small \small z'(x)=e^{-x+x^3}\cdot (-1+3x^{2})$

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Example Question #456 : Ap Calculus Ab

Given the function $\small \small \small \small \small z(x)=e^{\cos x+\sin x}$ , find its derivative.

Possible Answers:

$\small \small \small z'(x)=e^{\cos x-\sin x}\cdot (-\sin x+\cos x)$

$\small \small \small z'(x)=e^{\cos x+\sin x}$

$\small \small \small z'(x)=e^{\cos x+\sin x}\cdot (\sin x+\cos x)$

$\small \small z'(x)=e^{\cos x+\sin x}\cdot (-\sin x+\cos x)$

Correct answer:

$\small \small z'(x)=e^{\cos x+\sin x}\cdot (-\sin x+\cos x)$

Explanation:

Given the function $\small \small \small \small \small z(x)=e^{\cos x+\sin x}$ , we can find its derivative using the chain rule, which states that

$\small [f(g(x))]'=f'(g(x))g'(x)$

where $\small \small \small \small f(x)=e^{x}$ and $\small \small \small \small \small \small g(x)=\cos x+\sin x$ for $\small z(x)$ . We have $\small \small f'(x)=e^x$ and $\small \small \small \small \small \small g'(x)=-\sin x+\cos x$ , which gives us

$\small \small z'(x)=e^{\cos x+\sin x}\cdot (-\sin x+\cos x)$

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Example Question #457 : Ap Calculus Ab

Given the function $\small \small \small \small z(x)=e^{e^x-x^5}$ , find its derivative.

Possible Answers:

$\small \small \small \small \small \small \small z'(x)=e^{e^x-x^5}\cdot (e^x-5x^4)$

$\small \small \small \small \small \small \small \small z'(x)=e^{e^x-5x^4}\cdot (e^x-x^5)$

$\small \small \small \small \small \small \small z'(x)=e^{e^x-5x^4}$

$\small \small \small \small \small \small \small z'(x)=e^{e^x-5x^4}\cdot (e^x-5x^4)$

Correct answer:

$\small \small \small \small \small \small \small z'(x)=e^{e^x-x^5}\cdot (e^x-5x^4)$

Explanation:

Given the function $\small \small \small \small z(x)=e^{e^x-x^5}$ , we can find its derivative using the chain rule, which states that

$\small [f(g(x))]'=f'(g(x))g'(x)$

where $\small \small \small \small f(x)=e^{x}$ and $\small \small \small \small \small g(x)=e^x-x^5$ for $\small z(x)$ . We have $\small \small f'(x)=e^x$ and $\small \small \small \small g'(x)=e^x-5x^4$ , which gives us

$\small \small \small \small \small \small \small z'(x)=e^{e^x-x^5}\cdot (e^x-5x^4)$

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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 #231 : Derivatives

Example Question #141 : Computation Of The Derivative

Example Question #446 : Ap Calculus Ab

Example Question #51 : Chain Rule And Implicit Differentiation

Example Question #141 : Computation Of The Derivative

Example Question #453 : Ap Calculus Ab

Example Question #454 : Ap Calculus Ab

Example Question #455 : Ap Calculus Ab

Example Question #456 : Ap Calculus Ab

Example Question #457 : Ap Calculus Ab

All AP Calculus AB Resources

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