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Calculus 2 : Derivatives

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Example Questions

Example Question #371 : Ap Calculus Bc

Find the derivative of: y=cos^-^1(x)+cos(x)

Possible Answers:

$\frac{1-sin(x)}{\sqrt{1+x^2}}$

$-\frac{1+sin(x)\sqrt{1-x^2}}{\sqrt{1-x^2}}$

$\frac{1+cos(x)}{\sqrt{1+x^2}}$

$-\frac{1+sin(x)\sqrt{1+x^2}}{\sqrt{1+x^2}}$

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

Correct answer:

$-\frac{1+sin(x)\sqrt{1-x^2}}{\sqrt{1-x^2}}$

Explanation:

The derivative of inverse cosine is:

$\frac{d}{dx} cos^-^1(x) = -\frac{1}{\sqrt{1-x^2}}$

The derivative of cosine is:

$\frac{d}{dx} cos(x)=-sin(x)$

Combine the two terms into one term.

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

$-\frac{1}{\sqrt{1-x^2}}-\frac{sin(x)\sqrt{1-x^2}}{\sqrt{1-x^2}}= -\frac{1+sin(x)\sqrt{1-x^2}}{\sqrt{1-x^2}}$

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Example Question #3 : Other Derivative Review

The position function of a car is modeled as s(t)=4x^2 + 8x +1 .

Find the speed of the car at t=2 . HINT: The derivative of the position function is the speed function.

Possible Answers:

Correct answer:

Explanation:

As hinted in the problem statement, we need to find the derivative.

Using the simple power rule for derivatives $f(x)=x^n \rightarrow f'(x)=nx^{n-1}$ we find the derivative to be,

s'(t)=v(t)=8x+8 .

Now, plug in t=2 to solve.

v(2)=16+8=24

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Example Question #4 : Other Derivative Review

$f(x)=e^{3x}sin\left(\frac{x}{3}\right)$

Calculate f'(x) .

Possible Answers:

$3e^{3x}cos\left(\frac{x}{3}\right)+\frac{1}{3}e^{3x}sin\left(\frac{x}{3}\right)$

$3e^{3x}sin{\frac{x}{3}}+\frac{1}{3}e^{3x}cos\left(\frac{x}{3}\right)$

$e^{3x}sin\left(\frac{x}{3}\right)+e^{3x}cos\left(\frac{x}{3}\right)$

$3e^{3x}sin\left(\frac{x}{3}\right)+e^{3x}cos\left(\frac{x}{3}\right)$

$e^{3x}cos\left(\frac{x}{3}\right)$

Correct answer:

$3e^{3x}sin{\frac{x}{3}}+\frac{1}{3}e^{3x}cos\left(\frac{x}{3}\right)$

Explanation:

To calculate this derivate we need to use the product rule, as we have two functions multiplied by each other.

The product rule is defined as

$\frac{d}{dx}f(x)\cdot g(x)=f'(x)\cdot g(x) + f(x)\cdot g'(x)$

Let's make $e^{3x}$ be our f(x) and $sin\left(\frac{x}{3}\right)$ be our g(x) .

We get

$3e^{3x}sin{\frac{x}{3}}+\frac{1}{3}e^{3x}cos\left(\frac{x}{3}\right)$ .

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Example Question #3 : Other Derivative Review

$y=2x^{e^{x}}$ . Find ${\frac{\mathrm{d} y}{\mathrm{d} x}}$

Possible Answers:

$\frac{\mathrm{d} y}{\mathrm{d} x}=2x^{e^{x}}\cdot e^{x}(\frac{1}{x}+\ln\left | x \right |)$

$\frac{\mathrm{d} y}{\mathrm{d} x}=x^{e^{x}}\cdot e^{x}(\frac{1}{x}-\ln\left | x \right |)$

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

$\frac{\mathrm{d} y}{\mathrm{d} x}=2x^{e^{x}}e^{x}\cdot \ln\left | x \right |$

Correct answer:

$\frac{\mathrm{d} y}{\mathrm{d} x}=2x^{e^{x}}\cdot e^{x}(\frac{1}{x}+\ln\left | x \right |)$

Explanation:

$x^{e^{x}}$ is a variable raised to a variable power. There is no no way to find ${\frac{\mathrm{d} y}{\mathrm{d} x}}$ explicityly as it follows no derivative formula. So we must start by rearranging the original equation, $y=2x^{e^{x}}$ , to a point that we can implicitly differentiate.

First isolate the $x^{e^{x}}$ , by dividing both sides by .

$\frac{y}{2}=x^{e^{x}}$

Now take the natural log of both sides of the equation.

$\ln\left |\frac{y}{2} \right |=\ln\left |x^{e^{x}} \right |$ .

This enables the use of log properties, specifically the property $\ln (m^{n})=n\cdot \ln(m)$ . Applying this property to the right side of the equation gives the following.

$\ln\left |\frac{y}{2} \right |=e^{x}\cdot \ln\left |x\right |$

Now that there isn't a variable raised to a variable power anymore, so we can differentiate implicitly without issue.

The left side follows the pattern, $\frac{\mathrm{d} }{\mathrm{d} u}(\ln\left | u \right |)=\frac{du}{u}$ .We use the product rule on the right hand side of the equation.

$\frac{\mathrm{d} }{\mathrm{d} x}(\ln\left |\frac{y}{2} \right |)=\frac{\mathrm{d} }{\mathrm{d} x}(e^{x}\cdot \ln\left |x\right |)$

$\frac{\frac{1}{2}\frac{\mathrm{d} y}{\mathrm{d} x}}{\frac{y}{2}}=(e^{x})(\frac{1}{x})+(e^{x})(\ln\left | x \right |)$

Now solve for ${\frac{\mathrm{d} y}{\mathrm{d} x}}$ . First, simplify the left side of the equation, and pull the greatest common factor out of the right side.

$\frac{\frac{\mathrm{d} y}{\mathrm{d} x}}y=e^{x}(\frac{1}{x}+\ln\left | x \right |)$

Now multiply both sides by to cancel it off the right side.

$\frac{\mathrm{d} y}{\mathrm{d} x}=y\cdot e^{x}(\frac{1}{x}+\ln\left | x \right |)$

Next, using the original equation, $y=2x^{e^{x}}$ , replace with $2x^{e^{x}}$ . this puts everything back in terms of x.

$\frac{\mathrm{d} y}{\mathrm{d} x}=2x^{e^{x}}\cdot e^{x}(\frac{1}{x}+\ln\left | x \right |)$

This is the final answer.

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Example Question #6 : Other Derivative Review

Differentiate: y=cot(3x)

Possible Answers:

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

-cos^2(3x)

-3cos^2(3x)

-csc^2(3x)

-3csc^2(3x)

Correct answer:

-3csc^2(3x)

Explanation:

In order to differentiate cotangent, write the rule for the derivative.

$\frac{d}{dx} cot(x) = -csc^2(x)$

We will also need to use chain rule and multiply the derivative of the inner function.

$\frac{d}{dx} cot(3x) = -csc^2(3x) \cdot 3$

The answer is: -3csc^2(3x)

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Example Question #2 : Rules Of Basic Functions: Power, Exponential Rule, Logarithmic, Trigonometric, And Inverse Trigonometric

Find the derivative of the function $y = \int_{0}^{x^2} e^{-t^2}dt$

Possible Answers:

None of the other answers.

x^2

$e^{-x^2}$

$2xe^{-x^4}$

Correct answer:

$2xe^{-x^4}$

Explanation:

We can use the (first part of) the Fundemental Theorem of Calculus to "cancel out" the integral.

$y = \int_{0}^{x^2} e^{-t^2}dt$ . Start

$y' = \frac{d}{dx}\int_{0}^{x^2} e^{-t^2}dt$ . Take the derivative of both sides with respect to .

To "cancel out" the integral and the derivative sign, verify that the lower bound on the integral is a constant (It's in this case), and that the upper limit of the integral is a function of , (it's x^2 in this case).

Afterward, plug x^2 in for , and ultilize the Chain Rule to complete using the Fundemental Theorem of Calculus.

$y' = e^{-(x^2)^2}(x^2)' = 2xe^{-x^4}$ .

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Example Question #5 : Other Derivative Review

Find the derivative of $y = \int^{x}_{0} \sin^{-1}(\sqrt{t})dt$ .

Possible Answers:

Does not exist

None of the other answers

$\frac{1}{\sqrt{1-x^2}}$

$\sin^{-1}(\sqrt{x})$

Correct answer:

$\sin^{-1}(\sqrt{x})$

Explanation:

Use the Fundemental Theorem of Calculus to find .

$y = \int^{x}_{0} \sin^{-1}(\sqrt{t})dt$ . Start

$y' = \frac{d}{dx}\int^{x}_{0} \sin^{-1}(\sqrt{t})dt$ . Take derivatives of both sides with respect to .

Use the Fundemental Theorem of Calculus

$\frac{d}{dx}\int^{f(x)}_{x_0} g(t) dt = g(f(x))\times f'(x)$

with $f(x) = x, g(t) = \sin^{-1}(\sqrt{t})$ , x_0 = 0 to obtain

$y' = \sin^{-1}(\sqrt{x})$ .

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Example Question #2 : Rules Of Basic Functions: Power, Exponential Rule, Logarithmic, Trigonometric, And Inverse Trigonometric

Find the derivative of the function $y= \int_{1}^{x}t^2e^t\ln(t)\sin(t)dt$

Possible Answers:

$\frac{2xe^x\cos{x}}{x}$

None of the other answers

Does not exist

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

Correct answer:

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

Explanation:

To find the derivative of this function, we need to use the Fundemental Theorem of Calculus Part 1 (As opposed to the 2nd part, which is what's usually used to evaluate definite integrals)

$y= \int_{1}^{x}t^2e^t\ln(t)\sin(t)dt$ . Start

$y'= \frac{d}{dx}\int_{1}^{x}t^2e^t\ln(t)\sin(t)dt$ . Take derivatives of both sides.

$y'= x^2e^x\ln(x)\sin(x)$ . "Cancel" the integral and the derivative. (Make sure that the upper bound on the integral is a function of , and that the lower bound is a constant before you cancel, otherwise you may need to use some manipulation of the bounds to make it so.)

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Example Question #11 : Other Derivative Review

Let f(x)=cos^3(x+1) . Determine the value of $f'(\pi)$ .

Possible Answers:

$-3cos^2(\pi+1)sin(\pi+1)$

$-3cos^2(\pi+1)$

$3cos^2(\pi+1)$

$3cos^2(\pi+1)sin(\pi+1)$

Correct answer:

$-3cos^2(\pi+1)sin(\pi+1)$

Explanation:

Let us first determine the derivative of f(x) by successive applications of the chain rule. The given function is a composition of three simpler functions: first add to , then take the cosine of the sum, and then cube the result. By the chain rule, we must differentiate with respect to each of these functions and take their product to obtain the derivative of the original function, as shown below:

f(x)=cos^3(x+1),

f'(x)=(3cos^2(x+1))(-sin(x+1))(1)

= -3cos^2(x+1)sin(x+1)

To determine the value of this function at $x=\pi$ , simply substitute $\pi$ for :

$-3cos^2(\pi+1)sin(\pi+1)$

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Example Question #12 : Other Derivative Review

Evaluate.

$\int \ \cos(2x ) \ dx$

Possible Answers:

$F(x) = \frac{\sin (2x)}{2}$

$F(x) = \frac{\cos (2x)}{2}$

$F(x) = - \frac{\sin (x)}{2}$

Answer not listed

$F(x) = - \frac{\cos (x)}{2}$

Correct answer:

$F(x) = \frac{\sin (2x)}{2}$

Explanation:

$\int_{a}^{b} f(x) dx$

In order to evaluate an integral, first find the antiderivative of f(x).

If then
If then $F(x) = \frac{x^{a+1}}{a+1} + C$
If $f(x) = \frac{1}{x}$ then
If then
If $f(x) = \cos(x)$ then $F(x) = \sin(x) + C$
If $f(x) = \sin(x)$ then $F(x) = - \cos(x) + C$
If $f(x) = \sec^2 (x)$ then $F(x) = \tan(x) + C$
Substitution rule $\int f (g(x)) g'(x) \ dx = \int f(u) \ du$ where

In this case, $f(x) = \cos(2x)$ .

The antiderivative is $F(x) = \frac{\sin (2x)}{2}$ .

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Calculus 2 : Derivatives

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #371 : Ap Calculus Bc

Example Question #3 : Other Derivative Review

Example Question #4 : Other Derivative Review

Example Question #3 : Other Derivative Review

Example Question #6 : Other Derivative Review

Example Question #2 : Rules Of Basic Functions: Power, Exponential Rule, Logarithmic, Trigonometric, And Inverse Trigonometric

Example Question #5 : Other Derivative Review

Example Question #2 : Rules Of Basic Functions: Power, Exponential Rule, Logarithmic, Trigonometric, And Inverse Trigonometric

Example Question #11 : Other Derivative Review

Example Question #12 : Other Derivative Review

All Calculus 2 Resources

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