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AP Calculus AB : Derivatives

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Example Questions

Example Question #101 : Chain Rule And Implicit Differentiation

Find the derivative of $\ln x^5$ .

Possible Answers:

$\frac{5x^4}{x}$

$5 \ln x^4$

$\frac{1}{x}$

$\frac{5}{x}$

$\frac{1}{x^5}$

Correct answer:

$\frac{5}{x}$

Explanation:

There are two ways to do this problem.

Method 1: Use the power rule to turn $\ln x^5$ into $5\ln x$ , the derivative of which is $\tfrac{5}{x}$ .

Method 2: Use the chain rule.

$(\ln x^5)'=(\tfrac{1}{x^5})(5x^4)=\tfrac{5}{x}$

Either way, the answer is $\frac{5}{x}$ .

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Example Question #101 : Chain Rule And Implicit Differentiation

Find the derivative of $e^{x^2}$ .

Possible Answers:

$xe^{x^2}$

$2xe^{x^2}$

2e^x

$2e^{x^2}$

$e^{x^2}$

Correct answer:

$2xe^{x^2}$

Explanation:

Using the chain rule (f(g))'=f'(g)*g' :

$(e^{x^2})'=(e^{x^2})(2x)=2xe^{x^2}$

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Example Question #192 : Computation Of The Derivative

Find the derivative of $\tan(xe^x)$ .

Possible Answers:

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

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

$(xe^x+e^x)\sec^2 (xe^x)$

$e^x\sec^2 (xe^x)$

$\sec^2 (xe^x)$

Correct answer:

$(xe^x+e^x)\sec^2 (xe^x)$

Explanation:

First, use the chain rule (f(g))'=f'(g)*g' :

$(\tan(xe^x))'=sec^2(xe^x)*(xe^x)'$

Now we have to use the product rule (fg)'=fg'+gf'

(xe^x)'=xe^x+e^x(1)

Our final answer:

$(xe^x+e^x)\sec^2 (xe^x)$

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Example Question #104 : Chain Rule And Implicit Differentiation

Find the derivative of $\tan^2(\sin x)$ .

Possible Answers:

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

$\tan(\sin x)\sec(\sin x)$

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

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

$\sec^4(\sin x)\cos x$

Correct answer:

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

Explanation:

This problem requires us to use the chain rule twice.

First, take the derivative of $(\tan x)^2$ .

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

Now substitute $\sin x$ for and multiply by the derivative of $\sin x$ .

$2\tan(\sin x)\sec^2(\sin x)\cos x$ is our final answer.

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Example Question #103 : Chain Rule And Implicit Differentiation

Given yx^2 + 2y + x = 10 , calculate the slope of the tangent line when x = 1 .

Possible Answers:

$-\frac{7}{3}$

$-\frac{17}{9}$

Correct answer:

$-\frac{7}{3}$

Explanation:

The slope of the tangent line is found by taking the derivative. In this case, because we don't have the function defined explicitly, we'll have to use implicit differentiation to get

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

Where the first two terms come from product rule. Factoring out a dy/dx and moving the other terms to the right, we get

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

And solving for dy/dx, we get

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

As the problem doesn't give us a y value, we'll need to go back to the original equation to find it. Plugging in x = 1, we have

y(1)^2 + 2y + 1 = 10

And simplifying yields

3y = 9 ; y = 3

Plugging the point (1, 3) into our expression for dy/dx, we have

$\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{-2(1)(3) - 1}{1^2 + 2} = -\frac{7}{3}$

Alternatively, this expression could have been solved for explicitly by factoring a y term out. Note,

yx^2 + 2y + x = y(x^2 + 2) + x = 10

And thus, $y = \frac{10-x}{x^2 + 2}$ . Using quotient rule, we know the derivative of this

is $\frac{-x^2 - 2 - 2x(10-x)}{(x^2 + 2)^2}$ .

Evaluated at x=1, this becomes

$\frac{-(1)^2 - 2 - 2(10-1)}{((1)^2 + 2)^2} = -\frac{21}{9} = -\frac{7}{3}$

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Example Question #104 : Chain Rule And Implicit Differentiation

For the equation, $x \sin(y) + y\cos(y) = 0$ , find $\frac{\mathrm{d} y}{\mathrm{d} x}$ .

Possible Answers:

$\frac{\mathrm{d} y}{\mathrm{d} x}=\cot(y)$

$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{-\sin(y)}{x\cos(y)-y\sin(y)+\cos(y)}$

$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{1}{x\cos(y)+\cos(y)+y}$

$\frac{\mathrm{d} y}{\mathrm{d} x}= y\tan(y)-\tan(y) -x$

Correct answer:

$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{-\sin(y)}{x\cos(y)-y\sin(y)+\cos(y)}$

Explanation:

Finding $\frac{\mathrm{d} y}{\mathrm{d} x}$ for an equation of this complexity requires implicit differentiation.

We need to take the derivative of both sides with respect to x.

For $x\sin(y)$ , we will apply the product rule and chain rule.

The product rule is as follows: $\frac{\mathrm{d} }{\mathrm{d} x}(f\cdot g)= {\color{Red} f}\cdot{\color{Green} g'} + {\color{Green} g}\cdot{\color{Red} f'}$ , where f and g are the two factors of the expression. In this case, and will be and $\sin(y)$ respectively. When we find ${\color{Green} g'}$ , we will use the chain rule. This is because the inside of the $\sin$ contains variables other than x. Recall that the derivative of with respect to is $\frac{\mathrm{d} y}{\mathrm{d} x}$ .

This all means that ${\color{Green} g'}=\cos(y)\cdot(\frac{\mathrm{d} y}{\mathrm{d} x})$

Observe that ${\color{Red} f'}=\frac{\mathrm{d} }{\mathrm{d} x}(x)$ which is just 1.

Applying all this we find that the derivative of $x\sin(y)$ is

${\color{Red} x}\cdot{\color{Green} \cos(y)\cdot(\frac{\mathrm{d} y}{\mathrm{d} x})} + {\color{Green} \sin(y)}\cdot{\color{Red} (1)}$ .

Now we will find the derivative of the next term, $y \cos(y)$ . This will also require product rule and chain rule. For this product rule, ${\color{Red} f}$ and ${\color{Green} g}$ will be ${\color{Red} y}$ and ${\color{Green} \cos(y)}$ respectively.

${\color{Red} f'}$ will be the derivative of with respect to , which is ${\color{Red} \frac{\mathrm{d} y}{\mathrm{d} x}}$ .

${\color{Green} g'}$ will require the chain rule. The derivative of $\cos$ is $-\sin$ , and the derivative of with respect to is $\frac{\mathrm{d} y}{\mathrm{d} x}$ .

This means that ${\color{Green} g' }= {\color{Green} -\sin(y)(\frac{\mathrm{d} y}{\mathrm{d} x})}$ .

Assembling the power rule gives the derivative of the second term as

${\color{Red} y}\cdot{\color{Green} (-\sin(y))(\frac{\mathrm{d} y}{\mathrm{d} x})} +{\color{Green} \cos(y)}\cdot{\color{Red} \frac{\mathrm{d} y}{\mathrm{d} x}}$

For the right hand side of the equation, the derivative of is still .

Putting all three derivatives together into one equation yields the following:

$x\cos(y)(\frac{\mathrm{d} y}{\mathrm{d} x}) + \sin(y) -y\sin(y)(\frac{\mathrm{d} }{\mathrm{d} x}) + \cos(y)(\frac{\mathrm{d} y}{\mathrm{d} x})=0$

The next step in implicit differentiation is solving for $\frac{\mathrm{d} y}{\mathrm{d} x}$ . To do this we move all terms that have no $\frac{\mathrm{d} y}{\mathrm{d} x}$ factor to one side, and all terms with a $\frac{\mathrm{d} y}{\mathrm{d} x}$ factor to the other side.

In this case, we already have all $\frac{\mathrm{d} y}{\mathrm{d} x}$ terms on the left, so we will move the non- $\frac{\mathrm{d} y}{\mathrm{d} x}$ terms to the right side.

$x\cos(y)(\frac{\mathrm{d} y}{\mathrm{d} x}) -y\sin(y)(\frac{\mathrm{d} }{\mathrm{d} x}) + \cos(y)(\frac{\mathrm{d} y}{\mathrm{d} x})=-\sin(y)$

Now we can pull the common factor of $\frac{\mathrm{d} y}{\mathrm{d} x}$ out of the left side. We will then divide by what is remaining.

$\frac{\mathrm{d} y}{\mathrm{d} x}[x\cos(y)-y\sin(y)+\cos(y)]=-\sin(y)$

$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{-\sin(y)}{x\cos(y)-y\sin(y)+\cos(y)}$

This is the final answer. It cannot be simplified.

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Example Question #201 : Computation Of The Derivative

Let $f(x)=[e^{2x}+\sin(e^{x^3})]^3$ . Find the derivative, f'(x) .

Possible Answers:

$f'(x)=6e^{6x}+9x^2e^{x^3}\sin^2(e^{x^3})\cos(e^{x^3})$

$f'(x)=[2e^{2x}+3x^2e^{x^3}\cos(e^{x^3})]^3$

$f'(x)=3[e^{2x}+\sin(e^{x^3})]^2[e^{2x}+\cos(e^{x^3})]$

$f'(x)=3[e^{2x}+\sin(e^{x^3})]^2[2e^{2x}+3x^2e^{x^3}\cos(e^{x^3})]$

Correct answer:

$f'(x)=3[e^{2x}+\sin(e^{x^3})]^2[2e^{2x}+3x^2e^{x^3}\cos(e^{x^3})]$

Explanation:

The shortest and simplest way to find the derivative of this function is to use the Chain Rule. The Chain Rule definition is $[f(g(x))]' = f'(g(x))\cdot g'(x)$ . This is somewhat difficult to read and work with at first. Putting it in words helps though. What this definition states is that the derivative of "layered functions" is the derivative of the outer function times the derivative of the inner function. When I say "layered functions", I mean functions inside other functions. In this problem, we have the function, $(e^{2x}+\sin(e^{x^3}))$ , inside of a cubic function , (u)^3 , where is holding the place of the inner function. The outer function is the cubic, while the inner function is the $(e^{2x}+\sin(e^{x^3}))$ .

Applying the chain rule to this pair of layers means applying the power rule to the outer function, then multiplying it by the derivative of the inner function. Doing so gives

$3(inner function)^2\cdot(inner function)'$

We will need to find the derivative of the inner function, $(e^{2x}+\sin(e^{x^3}))$ , but first we will write the expression using the actual inner function.

$3(e^{2x}+\sin(e^{x^3}))^2\cdot(e^{2x}+\sin(e^{x^3}))'$ .

To find $(e^{2x}+\sin(e^{x^3}))'$ , we will take the derivative of the two terms inside separately.

The derivative of $e^{2x}$ is $e^{2x}\cdot(2)$ .

The derivative of $\sin(e^{x^3})$ is another Chain Rule. We take the derivative of outer function, $\sin$ , to get $\cos$ of the same inner function. Then we multiply it by the derivative of the inner function. The derivative of $e^{x^3}$ is $e^{x^3}\cdot(3x^2)$ .

Putting these together we get the following for the derivative of $(e^{2x}+\sin(e^{x^3}))$ :

$e^{2x}(2)+ \cos(e^{x^3})(e^{x^3}(3x^2))$

Simplifying it, we get

$2e^{2x} + 3x^2e^{x^3}\cos(e^{x^3})$

Putting this at the end of the original chain rule we have

$f'(x)=3(e^{2x}+\sin(e^{x^3}))^2\cdot(2e^{2x}+ 3x^2e^{x^3}\cos(e^{x^3}))$

This cannot be simplified, so it is the final answer

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Example Question #101 : Chain Rule And Implicit Differentiation

Find f'(x) if $f(x)=sin(e^{x})$

Possible Answers:

e^xx(cos(x))

-e(cos(e^x))

e^x(cos(e^x))

$\frac{cos(e^x)}{ln(x)}$

-ln(cos(x))

Correct answer:

e^x(cos(e^x))

Explanation:

$\frac{d}{du}[sin(u)]=cos(u)$

$\frac{d}{du}[e^u]=e^u \cdot u'$

According to the chain rule $\frac{d}{du}[f(g(u))]=f'(g(u)) \cdot g'(u)$ .

Therefore, the derivative we are looking for will be

$cos(e^x) \cdot e^x \cdot 1 = e^x(cos(e^x))$

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Example Question #102 : Chain Rule And Implicit Differentiation

Use implicit differentiation to calculate $\frac{\mathrm{d}y }{\mathrm{d} x}$ for the following equation: $e^{x}-cos(y)=x$

Possible Answers:

$(1+e^{x})(sec(y)$

$(1+e^{x})(csc(y)$

$(1-e^{x})(sec(y)$

$(1-e^{x})(csc(y))$

Correct answer:

$(1-e^{x})(csc(y))$

Explanation:

$e^{x}-cos(y)=x$

Differentiate both sides of the equation: $\frac{\mathrm{d} }{\mathrm{d} x}(e^{x}-cos(y)=x)$

Simplify: $e^{x}-\frac{\mathrm{d} }{\mathrm{d} x}cos(y)=1$

Use implicit differentiation to evaluate $\frac{\mathrm{d} }{\mathrm{d} x}cos(y)$ : $e^{x}-(-siny)\frac{\mathrm{d} y}{\mathrm{d} x}=1$

Simplify: $e^{x}+siny\frac{\mathrm{d} y}{\mathrm{d} x}=1$

Subtract $e^{x}$ from both sides of the equation: $siny\frac{\mathrm{d} y}{\mathrm{d} x}=1-e^{x}$

Divide both sides of the equation by siny: $\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{1-e^{x}}{sin(y)}$

Simplify: $\frac{\mathrm{d} y}{\mathrm{d} x}=1-e^{x}(csc(y))$

Solution: $\frac{\mathrm{d} y}{\mathrm{d} x}=1-e^{x}(csc(y))$

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Example Question #282 : Derivatives

What is the concavity of f(x)=6x^4-7x^3+8x^2-9x^2+18x at x=1?

Possible Answers:

Concave Diagonal

No Concavity

Concave Down

Concave Up

Concave Horizontal

Correct answer:

Concave Up

Explanation:

First, find the double derivative of the function given.

You should get f''(x)=72x^2-42x-2 .

Next, plug in x=1 to get f''(1)=28 .

The double derivative is a positive number, which means the concavity is Concave Up.

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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 #101 : Chain Rule And Implicit Differentiation

Example Question #101 : Chain Rule And Implicit Differentiation

Example Question #192 : Computation Of The Derivative

Example Question #104 : Chain Rule And Implicit Differentiation

Example Question #103 : Chain Rule And Implicit Differentiation

Example Question #104 : Chain Rule And Implicit Differentiation

Example Question #201 : Computation Of The Derivative

Example Question #101 : Chain Rule And Implicit Differentiation

Example Question #102 : Chain Rule And Implicit Differentiation

Example Question #282 : Derivatives

All AP Calculus AB Resources

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