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Calculus 2 : Derivative Review

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Example Questions

Example Question #1476 : Calculus Ii

Find the derivative of ln(sin(2x)).

Possible Answers:

2cot(2x)

tan(2x)

2tan(2x)

cot(2x)

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

Correct answer:

2cot(2x)

Explanation:

This is a chain rule derivative. For those problems, we need to start with the outermost derivative and work our way inwards. The very outermost function is ln(x). The derivative of that function is $\frac{1}{x}.$ Here, we simply replace with sin(2x). Then, we need to multiply by the derivative of the next chain, The derivative of that function is 2cos(2x). Putting it all together:

$ln(sin(2x))'=\frac{1}{sin(2x)}2cos(2x)=\frac{2cos(2x)}{sin(2x)}=2cot(2x).$

In the last step, we take advantage that $cotx=\frac{cosx}{sinx}.$

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Example Question #351 : Derivatives

Find the derivative of the following function:

e^x(3+e^x) .

Possible Answers:

$3e^x+e^{x^2}$

3+e^x

$3e^x+2e^{2x}$

5e^x

$3e^x+2xe^{x^2}$

Correct answer:

$3e^x+2e^{2x}$

Explanation:

First, we need to simplify the problem by distributing through the parenthesis.

$e^x(3+e^x)=3e^x+e^{2x}.$

Remember, for the second term, if we add exponents when multiplying functions with a common base.

Now, let's take the derivative! The function e^x returns itself, and the second term is a chain rule application.

$(3e^x+e^{2x})'=3e^x+2e^{2x}.$

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Example Question #351 : Derivative Review

Find the derivative of the following function:

4x^4-3x^3+2x^2-x .

Possible Answers:

12x^3-6x^2+4x-1

16x^3-9x^2+4x-1

16x^3-6x^2+4x-1

16x^3-9x^2+4x

12x^3-9x^2+4x

Correct answer:

16x^3-9x^2+4x-1

Explanation:

For polynomial derivatives, we use the power rule. We move the exponent to front of the function (and multiply it by the existing coefficient). Then, we reduce each exponent by one.

(4x^4-3x^3+2x^2-x)'=

$4*4x^{4-1}-3*3x^{3-1}+2*2x^{2-1}-1*x^{1-1}. =$

16x^3-9x^2+4x-1.

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Example Question #151 : First And Second Derivatives Of Functions

Find the first and second derivatives of the function,

$y=\ln(x^2 +1)$

Possible Answers:

First Derivative

$\frac{dy}{dx}=\frac{-x}{x^2+1}$

Second Derivative

$\frac{d^2y}{dx^2}= \frac{2x^2+1}{(x^2+1)}$

First Derivative

$\frac{dy}{dx}=\frac{2x}{x^2+1}$

Second Derivative

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

First Derivative

$\frac{dy}{dx}=\frac{2x}{x^2+1}$

Second Derivative

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

First Derivative

$\frac{dy}{dx}=\frac{x}{x^2+1}$

Second Derivative

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

First Derivative

$\frac{dy}{dx}=\frac{1}{x^2+1}$

Second Derivative

$\frac{d^2y}{dx^2}= \frac{2x}{(x^2+1)^2}$

Correct answer:

First Derivative

$\frac{dy}{dx}=\frac{2x}{x^2+1}$

Second Derivative

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

Explanation:

$y=\ln(x^2 +1)$

Finding the first derivative:

Recall the derivative of the natural logarithm function is,

$\frac{d}{du}\ln(u)=\frac{1}{u}$ . (1)

Proceed using equation (1) and the chain rule,

$\frac{dy}{dx}=\left(\frac{1}{x^2 +1}\right) \left(2x \right )=\frac{2x}{x^2+1}$

$\frac{dy}{dx}=\frac{2x}{x^2+1}$ (2)

To find the second derivative, use the quotient rule on equation (2).

$\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{2x}{x^2+1} \right )$ $=\frac{(x^2+1)\frac{d}{dx}(2x)-2x\frac{d}{dx}(x^2+1)}{(x^2+1)^2}$

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

$=\frac{2x^2+2-4x^2}{(x^2+1)^2}$

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

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

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Example Question #351 : Derivative Review

Determine the second derivative of a function k(x) , where the original function is expressed as 20x^3+30x^2-144x .

Possible Answers:

None

120x+60

120x-60

-2(60+x)

Correct answer:

120x+60

Explanation:

Step 1: Find the first derivative of the function:

By using the power rule which states, $\frac{d}{dx}x^n=nx^{n-1}$ , the first derivative is,

f'(x)=60x^2+60x-144

Step 2: Find the second derivative. To do this, just take the derivative of the former function by using the power rule again.

f''(x)=120x+60 .

Remember: The derivative of a constant is always zero!!

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Example Question #151 : First And Second Derivatives Of Functions

Find the first derivative of the following equation: $y=x^{sin^{2}x}$

Possible Answers:

$\frac{\mathrm{dy} }{\mathrm{d} x}=x^{sinx}(2sinxcosxlnx+\frac{sin^{2}x}{x})$

$\frac{\mathrm{dy} }{\mathrm{d} x}=x^{sin^{2}x}(2sinxcosxlnx+\frac{sin^{2}x}{x})$

$\frac{\mathrm{dy} }{\mathrm{d} x}=x(2sinxcosxlnx+\frac{sin^{2}x}{x})$

$\frac{\mathrm{dy} }{\mathrm{d} x}=x^{xsinx}(2sinxcosxlnx+\frac{sin^{2}x}{x})$

Correct answer:

$\frac{\mathrm{dy} }{\mathrm{d} x}=x^{sin^{2}x}(2sinxcosxlnx+\frac{sin^{2}x}{x})$

Explanation:

Take natural log of both sides: $lny=ln(x^{sin^{2}x})$

Using rules of logs and exponents, bring down $sin^{2}x$ , on right side of equation: $lny=sin^{2}xln(x)$

Differentiate each side: $\frac{\mathrm{d} }{\mathrm{d} x}lny=\frac{\mathrm{d} }{\mathrm{d} x}(sin^{2}xlnx)$

Use implicit differentiation on the left side of the equation, and the product rule on the right side: $\frac{1}{y}\frac{dy}{dx}=\frac{\mathrm{d} }{\mathrm{d} x}(sin^{2})lnx+sin^{2}x\frac{\mathrm{d} }{\mathrm{d} x}lnx$

Simplify: $\frac{1}{y}\frac{dy}{dx}=2sinxcosxlnx+\frac{sin^2{x}}{x}$

Multiply both sides of the equation by : $\frac{dy}{dx}=y(2sinxcosxlnx+\frac{sin^2{x}}{x})$

Simplify (remember that $y=x^{sin^{2}x}$ , from our original equation: $\frac{dy}{dx}=x^{sin^{2}x}(2sinxcosxlnx+\frac{sin^2{x}}{x})$

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Example Question #352 : Derivatives

What is the first derivative of f(x)=2x^2-7 ?

Possible Answers:

Correct answer:

Explanation:

Step 1: Use the power rule to simplify the exponent:

2x^2 becomes

Step 2: Take the derivative of the constant:

becomes because the derivative of any constant is zero.

The first derivative is the sum of everything that we have left after taking the derivative:

Final answer is

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Example Question #1481 : Calculus Ii

Find the first derivative of: $f(x)=3x^2+4x^{-1}-11$

Possible Answers:

$-6x^{-1}-4x^{2}$

$6x-4x^{-2}$

$6x+4x^{-2}$

$-6x-4x^{-2}$

Correct answer:

$6x-4x^{-2}$

Explanation:

Step 1: Use power rule on the first term:

3x^2 becomes .

Step 2: Use power rule on the second term:

$4x^{-1}$ becomes $-4x^{-2}$

Step 3: Take derivative of the third term:

becomes .

Step 4: Take answers from Steps 1-3...

Final derivative: $6x-4x^{-2}$

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Example Question #1483 : Calculus Ii

What is the first derivative of 2x-x^3 ?

Possible Answers:

2-3x^2

-2+3x^2

2+3x^2

-2-3x^2

Correct answer:

2-3x^2

Explanation:

Step 1: Use the power rule on the first term:

becomes

Step 2: Use the power rule on the second term:

-x^3 becomes -3x^2

Step 3: Add the final results for step 1 and step 2:

2-3x^2

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Example Question #351 : Derivatives

What is the second derivative of 3x+2x^2-4x^7 ?

Possible Answers:

4-168x^5

-4-128x^5

3-4x+28x^6

4+128x^6

Correct answer:

4-168x^5

Explanation:

Step 1: Use the power rule and take the derivative...

(3x+2x^2-4x^7)'=3+4x-28x^6

Step 2: Use the power rule and take the derivative of the result...

(3+4x-28x^6)'=4-168x^5

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Calculus 2 : Derivative Review

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #1476 : Calculus Ii

Example Question #351 : Derivatives

Example Question #351 : Derivative Review

Example Question #151 : First And Second Derivatives Of Functions

Example Question #351 : Derivative Review

Example Question #151 : First And Second Derivatives Of Functions

Example Question #352 : Derivatives

Example Question #1481 : Calculus Ii

Example Question #1483 : Calculus Ii

Example Question #351 : Derivatives

All Calculus 2 Resources

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