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Calculus 2 : Derivatives

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9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1621 : Calculus Ii

Differentiate $f(y)=\frac{e^y}{y}$ .

Possible Answers:

e^y

$\frac{(y-1)e^y}{y^2}$

e^y(ln(y))

$\frac{e^y-ye^y}{y}$

$\frac{e^y-ye^y}{y^2}$

Correct answer:

$\frac{(y-1)e^y}{y^2}$

Explanation:

Using the Quotient rule for derivatives, we know that the derivative will equal : $\frac{y(e^y)-e^y(1)}{y^2}$ .

If you simplify this, you will get:

$\frac{(y-1)e^y}{y^2}$

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Example Question #1621 : Calculus Ii

Differentiate $\frac{2}{x}+\frac{1}{4x^3}-\frac{6}{x^5}$ .

Possible Answers:

2lnx-3ln(4x)-30lnx

$-2x^{-2}-\frac{1}{4}x^{-4}-30x^{-6}$

$-2x^{-2}+\frac{1}{4}x^{-4}+30x^{-6}$

$2+\frac{1}{8}x^{-2}-\frac{6}{4}x^{-4}$

$2-\frac{1}{8}x^{-2}+\frac{6}{4}x^{-4}$

Correct answer:

$-2x^{-2}-\frac{1}{4}x^{-4}-30x^{-6}$

Explanation:

Rewriting the original equation gives us $f(x)=2(x^{-1})+\frac{1}{4}x^{-3}-6(x^{-5})$ .

Then, one can just use the power rule to get: $-2x^{-2}-\frac{3}{12}x^{-4}-30x^{-6}$ .

Simplifying this gives us: $-2x^{-2}-\frac{1}{4}x^{-4}-30x^{-6}$ .

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Example Question #1621 : Calculus Ii

Express the derivative of $y(x)=\sqrt {7x^3-6x^2+9x}$ in simplest terms.

Possible Answers:

$\large y'(x)=\frac {7x^2-4x+3}{\sqrt [3] {(7x^3-6x^2+9x)^2}}$

$\large y'(x)=\frac {7x^2-4x+3}{\sqrt {(7x^3-6x^2+9x)^3}}$

$\large y(x)=\bigg[\frac {7x^2-4x+3}{\sqrt{(7x^3-6x^2+9x)}}\bigg]^{\frac {2}{3}}$

$\large y'(x)=-\frac {7x^2-4x+3}{\sqrt [3] {(7x^3-6x^2+9x)^2}}$

Correct answer:

$\large y'(x)=\frac {7x^2-4x+3}{\sqrt [3] {(7x^3-6x^2+9x)^2}}$

Explanation:

Screen shot 2016 01 01 at 10.25.54 pm

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Example Question #502 : Derivative Review

Find the derivative of $\large y(x)=(2x+x^{-3})^\frac {1}{4}$ in simplest form.

Possible Answers:

$\large \large y'(x)=\frac {x^\frac {-7}{4}(2x^4-3)}{4(2x^4+1)^\frac {3}{4}}$

$\large \large y'(x)=\frac {(2x^4-3)}{4x^\frac {7}{4}(2x^4+1)^\frac {3}{4}}$

$\large \large y'(x)=\frac {x^{\frac {1}{4}}(2x^4-3)}{4(2x^4+1)^\frac {-3}{4}}$

$\large \large y'(x)=\frac {5(2x^4-3)}{4x^\frac {7}{4}(2x^4+1)^\frac {3}{4}}$

Correct answer:

$\large \large y'(x)=\frac {(2x^4-3)}{4x^\frac {7}{4}(2x^4+1)^\frac {3}{4}}$

Explanation:

Screen shot 2016 01 02 at 8.27.30 am

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Example Question #1622 : Calculus Ii

What is the first derivative of $\large y(x)=tan^4\sqrt {cot(3x)}$ ?

Possible Answers:

$\large - {6csc^2(3x)\big (tan^3 \sqrt {cot(3x))}) \big (sec^2)}$

$\large \frac {6csc^2(3x)\big (tan^3 \sqrt {cot(3x))}) \big (sec^2 \sqrt {cot(3x)})}{\sqrt {cot(3x)}}$

$\large -\frac {6csc^2(3x)\big (tan^3 \sqrt {cot(3x))}) \big (sec^2 \sqrt {cot(3x)})}{\sqrt {cot(3x)}}$

$\large \large -\frac {6csc^2(3x)\big (tan \sqrt {cot(3x))}\big )^3 \big (sec^2 \sqrt {cot(3x)})}{\sqrt {cot(3x)}}$

Correct answer:

$\large -\frac {6csc^2(3x)\big (tan^3 \sqrt {cot(3x))}) \big (sec^2 \sqrt {cot(3x)})}{\sqrt {cot(3x)}}$

Explanation:

Screen shot 2016 01 02 at 9.36.43 am

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Example Question #21 : Other Derivative Review

Which of the following IS NOT a rule used when finding derivatives of any function?

Possible Answers:

Chain Rule

Exponential Rule of Logarithms

Power Rule

Product Rule

Correct answer:

Exponential Rule of Logarithms

Explanation:

Step 1: Recall any rules that are used in derivatives...

Power Rule
Quotient Rule
Product Rule
Chain Rule

Step 2: Look at the choices in the question and compare the ones listed in Step 1.

We see Power Rule, Chain Rule, and Product Rule.

There is no such rule called the Addition rule, so this is the incorrect answer.

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Example Question #502 : Derivatives

Suppose , , and any composite function between them are defined and differentiable everywhere. Given the derivatives for and :

h'(x)=2x

g'(x)=4

Find the derivative of f(x) where,

f(x)=h(x-g(x))+g(x)

Possible Answers:

f'(x)=-6x

f'(x)=6x+6g'(x)+4

2x+4

f'(x)=6g(x)+4

f'(x)=6x-6g(x)+4

Correct answer:

f'(x)=6x-6g(x)+4

Explanation:

f(x)=h(x-g(x))+g(x)

This is a conceptual problem. First notice that the function f(x) is the sum of two functions h(x-g(x)) and g(x) . We must differentiate each term.

To differentiate the first term, notice that h(x-g(x)) is a function of the function x-g(x) , so we must use the chain rule to differentiate with respect to . We cannot conclude that h'(x-g(x))=2x .

We were given h'(x)=2x . What this equation is telling us is that the function h(x) will have a derivative that is 2 times whatever is inside the parenthesis, with respect to the whatever is inside the parenthesis. If we wish to use to differentiate the composite h(x-g(x)) we can start with the derivative with respect to x-g(x) , which will be 2(x-g(x)) .

Now if we want to differentiate with respect to , we take the derivative with respect to x-g(x) and multiply by the derivative of x-g(x) with respect to (the chain rule).

$h'(x-g(x))=2(x-g(x))\times(1-g'(x))$

We were given g'(x)=4

$h'(x-g(x))=2(x-g(x))\times(-3)$

h'(x-g(x))= 6x-6g(x)

Now we can write the derivative of f(x) ,

f'(x)=h'(x-g(x))+g'(x)

f'(x)=6x-6g(x)+4

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Example Question #501 : Derivative Review

Find the equation of the tangent line to the curve,

$y=\ln(x)+3x^2$

at x = 2

Find the equation of the tangent line to the curve,

$y=\ln(x)+3x^2$

at x = 2

Possible Answers:

$y = -\frac{25}{2}x+\ln(2) +13$

$y = \frac{25}{2}x -12$

$y = \frac{25}{2}x+\ln(2) -13$

$y = -\frac{25}{2}x+10$

$y = \frac{25}{2}x+\ln(2)$

Correct answer:

$y = \frac{25}{2}x+\ln(2) -13$

Explanation:

Find the equation of the tangent line to the curve corresponding to

$f(x)=\ln(x)+3x^2$

The first step is to compute the derivative for the function and then evaluate the derivative at x = 2

$f'(x)=\frac{1}{x}+6x\: \: \: \:\: \: \:\Rightarrow \: \: \: \: \: \: \: \:f'(2)=\frac{1}{2}+6(2)=\frac{25}{2}$

Therefore $\frac{25}{2}$ will be the slope of the tangent line at x = 2 . Write an equation for the line,

y = mx+b

$y = \frac{25}{2}x+b$

Now we need to find the y-intercept. Use the original function to find when x=2 .

$f(2)=\ln(2)+3(2^2)=\ln(2)+12$

This gives us the point at which the tangent line meets the curve,

$(2\: \: ,\: \, \ln(2)+12)$

Now use this point to find the y-intercept,

$b = y-\frac{25}{2}x$

$b = \ln(2)+12-\frac{25}{2}(2)$

$b=\ln(2)-13$ .

The equation of the line is therefore,

$y = \frac{25}{2}x+\ln(2) -13$

Plot

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Example Question #502 : Derivative Review

Differentiate the function:

f(x)=xln(x)-x^2

Possible Answers:

f'(x)=ln(x)-2x-1

f'(x)=ln(x)-2x+1

f'(x)=ln(x)+2x+1

f'(x)=xln(x)+2x+1

f'(x)=ln(x)+2x-1

Correct answer:

f'(x)=ln(x)-2x+1

Explanation:

First we see from the sum rule that:

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

The first term we use the product rule to differentiate:

(fg)'=f'g+g'f

$(x\cdot ln(x))'=1\cdot ln(x)+\frac{x}{x}=ln(x)+1$

The second term is:

f'(x^2)=2x

Therefore:

f'(xln(x)-x^2)=ln(x)+1-2x

=ln(x)-2x+1

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Example Question #503 : Derivative Review

Differentiate the following function:

y=ln(cos(x))

Possible Answers:

y'=-tan(x)

$y'=-\frac{1}{sin(x)}$

$y'=\frac{1}{sin(x)}$

y'=tan(x)

$y'=\frac{1}{cos(x)}$

Correct answer:

y'=-tan(x)

Explanation:

To differentiate the function y=ln(cos(x)) we have to use the chain rule

$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$

let u=cos(x) therefore y=ln(u) and

$\frac{dy}{du}=\frac{1}{u}=\frac{1}{cos(x)}$

and

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

Therefore:

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

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Calculus 2 : Derivatives

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #1621 : Calculus Ii

Example Question #1621 : Calculus Ii

Example Question #1621 : Calculus Ii

Example Question #502 : Derivative Review

Example Question #1622 : Calculus Ii

Example Question #21 : Other Derivative Review

Example Question #502 : Derivatives

Example Question #501 : Derivative Review

Example Question #502 : Derivative Review

Example Question #503 : Derivative Review

All Calculus 2 Resources

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