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Calculus 2 : First and Second Derivatives of Functions

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Example Questions

Example Question #371 : Derivatives

Compute the first derivative of the following function:

$f(x)=4x^{-3}+6\sin(2x)+3e^{4x}$

Possible Answers:

$12(e^{4x}+\cos(2x)-x^{-2})$

$12e^{4x}+6\cos(2x)-12x^{-4}$

$12e^{4x}-6\cos(2x)-12x^{-4}$

$12(e^{4x}+\cos(2x)-x^{-4})$

$12(e^{4x}-\cos(2x)-x^{-2})$

Correct answer:

$12(e^{4x}+\cos(2x)-x^{-4})$

Explanation:

The first term of the function's derivative is found simply with the power rule:

$\frac{d}{dx}(x^n)=nx^{n-1}\rightarrow \frac{d}{dx}(4x^{-3})=-12x^{-4}$

The second term is a trig derivative with the chain rule:

$\frac{d}{dx}(6\sin(2x))=6\cos(2x)\cdot 2=12\cos(2x)$

The third term requires the rule for derivatives of exponential and the chain rule:

$\frac{d}{dx}(e^x)=e^x\rightarrow \frac{d}{dx}(3e^{4x})=3e^{4x}\cdot 4=12e^{4x}$

Combining these three terms, we get the final answer:

$12e^{4x}+12\cos(2x)-12x^{-4} =12(e^{4x}+\cos(2x)-x^{-4})$

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Example Question #1493 : Calculus Ii

Compute the first derivative of the following function using the product rule:

h(x)=(3x^2+7)(4x^3+x)

Possible Answers:

12x^5+31x^3+7x

60x^4+93x^2+7

36x^4+84x^2+7

48x^5+34x^3+42x

72x^3+6x

Correct answer:

60x^4+93x^2+7

Explanation:

The rule for the derivative of the product of two functions is as follows:

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

For the expression given, the two functions are:

$f(x)=3x^2+7, \: \: \:g(x)=4x^3+x$

Whose derivatives are:

$f'(x)=6x,\: \: \:g'(x)=12x^2+1$

The derivative of the product is then:

$\frac{d}{dx}[f(x)\cdot g(x)]=(6x)(4x^3+x)+(3x^2+7)(12x^2+1)$

Simplified, this reduces to:

24x^4+6x^2+36x^4+3x^2+84x^2+7=60x^4+93x^2+7

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Example Question #173 : First And Second Derivatives Of Functions

Compute the first derivative of the following function:

$f(x)=4\sin(2x)e^{3x}$

Possible Answers:

$24\cos(2x)e^{3x}+4\sin(2x)e^{3x}$

$8e^{3x}(3\sin(2x)+2\cos(2x))$

$8e^{3x}(2\sin(2x)+3\cos(2x))$

$24\cos(2x)e^{3x}$

$4e^{3x}(3\sin(2x)+2\cos(2x))$

Correct answer:

$4e^{3x}(3\sin(2x)+2\cos(2x))$

Explanation:

The rule for the derivative of the product of two functions is as follows:

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

For the expression given, the two functions are:

$f(x)=4\sin(2x), \: \: \:g(x)=e^{3x}$

Whose derivatives are:

$f'(x)=8\cos(2x),\: \: \:g'(x)=3e^{3x}$

The derivative of the product is then:

$\frac{d}{dx}[f(x)\cdot g(x)]=8\cos(2x)\cdot e^{3x}+4\sin(2x)\cdot 3e^{3x}$

Simplifying, we get:

$\frac{d}{dx}[f(x)\cdot g(x)]=8\cos(2x)\cdot e^{3x}+12\sin(2x)\cdot e^{3x} =4e^{3x}(2\cos(2x)+3\sin(2x))$

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Example Question #1494 : Calculus Ii

Find the first derivative of f(x)=2x^2+4 ?

Possible Answers:

$\frac {2x^3}{3}+4x$

Correct answer:

Explanation:

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

2x^2 becomes

Step 2: Take derivative of second term...

becomes

Note: The derivative of a(n) constant term is always zero.

The first derivative is

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Example Question #175 : First And Second Derivatives Of Functions

What is the first derivative of $\frac {1}{3x^2}$ ?

Possible Answers:

$-\frac {2}{3x^3}$

$\frac {1}{x^4}$

x^3

$\frac {2}{3x^3}$

Correct answer:

$-\frac {2}{3x^3}$

Explanation:

Step 1: Find the derivative of f(x) and g(x) , which we denote as f'(x) and g'(x) .

f'(x)=(1)'=0

g'(x)=(3x^2)'=6x

Step 2: Plug in the functions into the quotient rule formula:

$\frac {0(3x^2)-6x(1)}{(3x^2)^2}$

Step 3: Simplify...

$\frac {0-6x}{9x^4}=\frac {-6x}{9x^4}$

Step 4: Reduce:

$\frac {-6x}{9x^4}=-\frac {2}{3x^3}$

The first derivative is $-\frac {2}{3x^3}$

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Example Question #171 : First And Second Derivatives Of Functions

Find $\large f''(x)$ if f'(x)=2x^3-7x ?

Possible Answers:

-6x^2+7

6x^2+7

-6x^2-7

6x^2-7

Correct answer:

6x^2-7

Explanation:

Step 1: We just take the derivative of f'(x) , which is given to us in the question.

(2x^3-7x)'

Step 2: Use the power rule to take the derivative..

First Term: $2x^3\rightarrow 6x^2$
Second Term: $-7x\rightarrow-7$

Step 3: Take the final answers in Step 2 and add them together...

The second derivative, f''(x) , is 6x^2-7

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Example Question #171 : First And Second Derivatives Of Functions

Find the first derivative of: $\large y=\left [ (2x+1)^2(x+7)^3\right ]$

Possible Answers:

$\large \large \large \large \large 16x^4-266x^3+1428x^2-2267x+833$

$\large -16x^4+266x^3-1428x^2+2276x-833$

$\large \large 16x^4+266x^3+1428x^2+2276x+833$

$\large 0$

Correct answer:

$\large \large 16x^4+266x^3+1428x^2+2276x+833$

Explanation:

Screen shot 2016 01 02 at 11.17.54 am

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Example Question #172 : First And Second Derivatives Of Functions

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

Possible Answers:

$f{}'(x)=9x^2+x-1$

$f{}'(x)=9x^2+x$

$f{}'(x)=9x^2+3x$

$f{}'(x)=x^2+x$

$f{}'(x)=9x^2+x-4$

Correct answer:

$f{}'(x)=9x^2+x$

Explanation:

Recall that when taking the derivative, multiply the exponent by the coefficient in front of the x term and then subtract one from the exponent:

$f{}'(x)=9x^2+x$

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Example Question #174 : First And Second Derivatives Of Functions

What is the first derivative of $\large \large f(x)=\sqrt {x}$

Possible Answers:

$\large \frac {1}{2\sqrt{x}}$

$\large 0$

$\large \frac {1}{2x}$

$\large \frac {1}{\sqrt {x}}$

Correct answer:

$\large \frac {1}{2\sqrt{x}}$

Explanation:

Step 1: Re-write $\large f(x)$ as $\large x$ to a power:

$\large \sqrt{x}=x^\frac {1}{2}$
Step 2: Use the power rule:

$\large x^\frac {1}{2}=\frac {1}{2}x^{\frac {1}{2}-1}$

Step 3: Simplify the exponent:

$\large \frac {1}{2}x^{\frac {1}{2}}-1=\frac {1}{2}x^\frac {-1}{2}$

Step 4: Re-write $\large x^\frac {-1}{2}$ as a fraction with a positive exponent:

$\large x^\frac {-1}{2}=\frac {1}{x^\frac {1}{2}}=\frac {1}{\sqrt{x}}$
Step 5: Multiply the result in Step 4 with the coefficient $\large \frac {1}{2}$ in step 3:

$\large \frac {1}{2} \cdot \frac {1}{\sqrt{x}}=\frac {1}{2\sqrt{x}}$

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Example Question #173 : First And Second Derivatives Of Functions

$\begin{align*}&\text{Find the derivative of the function }\\&f(x)=-11cos(14x^{2})\end{align*}$

Possible Answers:

$308xsin(14x^{2})$

28x

$11sin(cos(14x^{2}))$

$-11cos(28xsin(14x^{2}))$

Correct answer:

$308xsin(14x^{2})$

Explanation:

$\begin{align*}&\text{The function }f(x)=-11cos(14x^{2})\text{, can be seen as a function nested within another.}\\&\text{Notice the }14x^{2}\text{ within the }-11cos()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[cos(u)]=-sin(u)du\\&u=14x^{2}\\&du=28x\\&f'(x)=308xsin(14x^{2})\end{align*}$

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Calculus 2 : First and Second Derivatives of Functions

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #371 : Derivatives

Example Question #1493 : Calculus Ii

Example Question #173 : First And Second Derivatives Of Functions

Example Question #1494 : Calculus Ii

Example Question #175 : First And Second Derivatives Of Functions

Example Question #171 : First And Second Derivatives Of Functions

Example Question #171 : First And Second Derivatives Of Functions

Example Question #172 : First And Second Derivatives Of Functions

Example Question #174 : First And Second Derivatives Of Functions

Example Question #173 : First And Second Derivatives Of Functions

All Calculus 2 Resources

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