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Calculus 2 : Other Derivative Review

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Example Question #1 : Other Derivative Review

Use implicit differentiation to find $\frac{dy}{dx}$ :

$x^4\sin(x+y)=2$

Possible Answers:

$\frac{dy}{dx} = -\left(\frac{4\sin(x+y)-2x}{x\cos x}\right)$

$\frac{dy}{dx} = -\left(4\tan(x+y) +1 \right )$

$\frac{dy}{dx} = 2-4x^3\sin (x+y)$

$\frac{dy}{dx} = \frac{4\tan(x+y)}{x}$

$\frac{dy}{dx}=-\left(\frac{4\tan(x+y)+x}{x} \right )$

Correct answer:

$\frac{dy}{dx}=-\left(\frac{4\tan(x+y)+x}{x} \right )$

Explanation:

To differentiate the left-hand side of the equation, we must use the product rule: f'(x)g(x) +f(x)g'(x) .

Let f(x)=x^4 so that f'(x)=4x^3 , and $g(x)=\sin(x+y)$ so that ${g'(x)=\cos(x+y) \cdot \left(1+\frac{dy}{dx} \right )}$ . After differentiation, we end up with ${4x^3\sin(x+y) +x^4\cos(x+y)\left(1+\frac{dy}{dx} \right )=0}$ .

Using algebra to isolate the $\frac{dy}{dx}$ term, we find that ${\frac{dy}{dx}= -\frac{4x^3 \sin(x+y)}{x^4\cos(x+y)}-1} =-\left(\frac{4\tan(x+y) +x}{x} \right )$ .

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Example Question #2 : Other Derivative Review

Find $\frac{dy}{dx}$ :

$y=x^{\sin x}$

Possible Answers:

$\frac{dy}{dx} = x^{\sin x}\left(\cos x \ln x + \frac{\sin x}{x} \right )$

$\frac{dy}{dx}= \ln\left( x^{\sin x}\right ) \cos x$

$\frac{dy}{dx} = -\cos x \left( x^{\sin x}\right)$

$\frac{dy}{dx} = \sin x\left(x^{\cos x} \right )$

$\frac{dy}{dx} = \cos x \ln x +\frac{\sin x }{x} \right$

Correct answer:

$\frac{dy}{dx} = x^{\sin x}\left(\cos x \ln x + \frac{\sin x}{x} \right )$

Explanation:

Since $\tiny{x}$ is a part of both the base and the exponent, we need to use logarithmic differentiation; that is, take the log of both sides of the equation: $\ln y = \ln (x^{\sin x}) \rightarrow \ln y =\sin x \ln x.$

Differentiating the latter equation, we obtain $\frac{1}{y}\frac{dy}{dx} = \cos x \cdot \ln x +\sin x \cdot \left(\frac{1}{x}\right)$ .

Thus, $\frac{dy}{dx} = x^{\sin x}\left(\cos x \ln x + \frac{\sin x}{x} \right )$ .

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Example Question #1 : Computation Of Derivatives

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

Possible Answers:

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

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

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

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

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

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 #5 : Other Derivative Review

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

Possible Answers:

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

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

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

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

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 #1 : Computation Of Derivatives

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

Possible Answers:

$e^{-x^2}$

$2xe^{-x^4}$

x^2

None of the other answers.

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 #7 : Other Derivative Review

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

Possible Answers:

Does not exist

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

None of the other answers

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

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 #1 : Other Derivative Review

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

Possible Answers:

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

Does not exist

None of the other answers

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

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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Calculus 2 : Other Derivative Review

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #1 : Other Derivative Review

Example Question #2 : Other Derivative Review

Example Question #1 : Computation Of Derivatives

Example Question #3 : Other Derivative Review

Example Question #4 : Other Derivative Review

Example Question #5 : Other Derivative Review

Example Question #6 : Other Derivative Review

Example Question #1 : Computation Of Derivatives

Example Question #7 : Other Derivative Review

Example Question #1 : Other Derivative Review

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