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Calculus 2 : Derivatives

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Example Questions

Example Question #61 : Derivative Review

Find the derivative of $f(x)=\frac{x}{x+1}$ using the definition of the derivative.

Possible Answers:

$f'(x)=\frac{1}{2x+1}$

f'(x)=1

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

$f'(x)=\frac{1}{5x^2-3}$

Correct answer:

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

Explanation:

The definition of a derivative is $f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f(x)=\frac{x}{x+1}$

$f(x+h)=\frac{x+h}{x+h+1}$

Substituting these expressions into the definition of a derivative gives us

$f'(x)=\lim_{h\rightarrow 0}\frac{1}{h}*(f(x+h)-f(x))$

$f'(x)=\lim_{h\rightarrow 0}\frac{1}{h}*\left (\frac{x+h}{x+h+1}-\frac{x}{x+1} \right )$

$f'(x)=\lim_{h\rightarrow 0}\frac{1}{h}*\left (\frac{(x+h)(x+1)-x(x+h+1)}{(x+h+1)(x+1))} \right )$

$f'(x)=\lim_{h\rightarrow 0}\left (\frac{(x^2+hx+x+h) -(x^2+hx+x)}{h(x^2+hx+x+x+h+1)} \right )$

$f'(x)=\lim_{h\rightarrow 0}\left (\frac{h}{h(x^2+hx+2x+h+1)} \right )$

$f'(x)=\lim_{h\rightarrow 0}\left (\frac{1}{x^2+hx+2x+h+1} \right )$

$f'(x)=\left (\frac{1}{x^2+(0)x+2x+(0)+1} \right )$

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

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Example Question #61 : Definition Of Derivative

Possible Answers:

x^2

4x+C

Correct answer:

Explanation:

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Example Question #61 : Derivatives

What is the derivative of a term: ax^2 ?

Possible Answers:

2ax

3x^2

2x^3

Correct answer:

2ax

Explanation:

Step 1: Move the exponent to the coefficient..

We get, 2ax..

Step 2: The new exponent after moving the old one down is less than it was before:

$2ax^{2-1}=2ax$

The derivative of ax^2 is 2ax

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Example Question #63 : Definition Of Derivative

Find the function represented by the following limit,

$\lim_{ h->0}\frac{\ln[(y+h)^2]-\ln(h^2)}{h}$

Possible Answers:

y^2 -1

$\frac{2}{y}$

$\frac{1}{y^2}+1$

Constant

$\ln(y+4)$

Correct answer:

$\frac{2}{y}$

Explanation:

$\lim_{ h->0}\frac{\ln[(y+h)^2]-\ln(h^2)}{h}$ (1)

Recall the first principles definition of a derivative,

$f'(y) = \lim_{h->0}\frac{f(y+h)-f(h)}{h}$ (2)

The limit in Equation (1) can be determined by inspection if we compare to the general definition of a derivative, Equation (2). It's apparent that equation (1) is the definition of the derivative for the function $\ln(y^2)$ . So the limit can be found by simply differentiating the function $\ln(y^2)$ with respect to .

$\frac{d}{dy}(\ln(y^2))=(2y)\frac{1}{y^2}=\frac{2}{y}$

$\lim_{ h->0}\frac{\ln[(y+h)^2]-\ln(h^2)}{h}=\frac{2}{y}$

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Example Question #61 : Definition Of Derivative

Find the derivative of the following function:

$f(x)=\frac{cosx}{sinx}$ .

Possible Answers:

sec^2x

secxtanx

-csc^2x

csc^2x

-sec^2x

Correct answer:

-csc^2x

Explanation:

The function in the problem is simply cotx ! Therefore, by knowing the derivative of cotx , we know the solution is -csc^2x . You would obtain the same results if you completed this problem using the quotient rule and simplifying.

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Example Question #1 : Derivative At A Point

$g(x) = \sec \frac{\pi}{x}$

Evaluate g '(6) .

Possible Answers:

$- \frac{\pi}{54}$

$- \frac{\pi}{9}$

$- \frac{\pi \sqrt{3}}{9}$

$- \frac{\pi \sqrt{3}}{27}$

$- \frac{\pi \sqrt{3}}{54}$

Correct answer:

$- \frac{\pi}{54}$

Explanation:

To find g'(x) , substitute $u =\frac{ \pi }{x} = \pi x ^{-1}$ and use the chain rule:

$g(x) = \sec \left (\frac{\pi}{x} \right )$

$g'(x) = \frac{\mathrm{d} }{\mathrm{d} x} \sec\frac{\pi}{x}$

$= \frac{\mathrm{d} }{\mathrm{d} x} \sec u$

$= \frac{\mathrm{d} u}{\mathrm{d} x} \cdot \frac{\mathrm{d} }{\mathrm{d}u} \sec u$

$= \frac{\mathrm{d} }{\mathrm{d} x} \pi x^{-1} \cdot \frac{\mathrm{d} } {\mathrm{d}u} \sec u$

$= \pi \cdot (-1) \cdot x^{-1-1} \cdot \tan u \sec u$

$= - \pi x^{-2} \cdot \tan \frac{\pi}{x} \sec \frac{\pi}{x}$

$= - \frac{\pi}{x^{2}} \tan \frac{\pi}{x} \sec \frac{\pi}{x}$

So $g '(x)= - \frac{\pi}{x^{2}}\tan \frac{\pi}{x} \sec \frac{\pi}{x}$

and

$g '(6)= - \frac{\pi}{6^{2}}\tan \frac{\pi}{6} \sec \frac{\pi}{6}$

$= - \frac{\pi}{36} \cdot \frac{\sqrt{3}}{3} \cdot \frac{2\sqrt{3}}{3}$

$= - \frac{\pi}{36} \cdot \frac{2\cdot 3}{9} = - \frac{\pi}{54}$

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Example Question #2 : Derivative At A Point

What is the equation of the line tangent to the graph of the function

$f(x)= 1+\frac{4}{x}$

at the point (2, 3) ?

Possible Answers:

y = x+1

y =2 x-1

y = 3

y =4 x-5

y = - x+5

Correct answer:

y = - x+5

Explanation:

The slope of the line tangent to the graph of $f(x)= 1+\frac{4}{x}$ at (2, 3) is

f'(2) , which can be evaluated as follows:

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

$f'(x) = (-1) 4x^{-1-1}$

$f'(x) = - \frac{4}{x^{2}}$

$f'(2) = - \frac{4}{2^{2}}= - 1$

The equation of the line with slope through (2, 3) is:

$y - y_{1} = m (x-x_{1})$

y - 3 = -1 (x-2)

y - 3 = - x+2

y = - x+5

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Example Question #2 : Derivative At A Point

What is the equation of the line tangent to the graph of the function

$f(x) = 2 \sin \frac{\pi x}{6}$

at the point (7, -1) ?

Possible Answers:

$y = - \frac {\pi \sqrt{3 }}{6} x+ \frac {7 \pi \sqrt{3 }}{6} -1$

$y = \frac {\pi \sqrt{2 }}{4} x- \frac {7 \pi \sqrt{2 }}{4} -1$

$y = \frac {\pi \sqrt{3 }}{6} x - \frac {7 \pi \sqrt{3 }}{6} -1$

$y = - \frac {\pi }{6} x+ \frac {7 \pi }{6} -1$

$y = - \frac {\pi \sqrt{2 }}{4} x+ \frac {7 \pi \sqrt{2 }}{4} -1$

Correct answer:

$y = - \frac {\pi \sqrt{3 }}{6} x+ \frac {7 \pi \sqrt{3 }}{6} -1$

Explanation:

The slope of the line tangent to the graph of $f(x) = 2 \sin \frac{\pi x}{6}$ at (7, -1) is

f'(7) , which can be evaluated as follows:

$f(x) = 2 \sin \frac{\pi x}{6} = 2 \sin \frac{\pi }{6}x$

$f'(x) = 2 \cdot\frac {\pi }{6} \cos \frac{\pi }{6}x$

$f'(x) = \frac {\pi }{3} \cos \frac{\pi }{6}x$

$f'(7) = \frac {\pi }{3} \cos \frac{7 \pi }{6}$

$= \frac {\pi }{3} \left ( - \frac{\sqrt{3}}{2}\right ) = - \frac {\pi \sqrt{3 }}{6}$ , the slope of the line.

The equation of the line with slope $- \frac {\pi \sqrt{3 }}{6}$ through (7, -1) is:

$y - y_{1} = m (x-x_{1})$

$y - (-1) = - \frac {\pi \sqrt{3 }}{6} (x-7)$

$y +1 = - \frac {\pi \sqrt{3 }}{6} x+\left ( \frac {\pi \sqrt{3 }}{6} \right ) 7$

$y = - \frac {\pi \sqrt{3 }}{6} x+ \frac {7 \pi \sqrt{3 }}{6} -1$

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Example Question #62 : Derivatives

What is the equation of the line tangent to the graph of the function

$f(x)= \frac{8}{x}$

at (-2,-4) ?

Possible Answers:

y = 2 x

y = -2 x-8

$y = \frac{1}{2}x-3$

y = -8 x-20

$y = - \frac{1}{2}x-5$

Correct answer:

y = -2 x-8

Explanation:

The slope of the line tangent to the graph of $f(x)= \frac{8}{x}$ at (-2,-4) is

f'(-2) , which can be evaluated as follows:

$f(x)= \frac{8}{x} = 8x^{-1}$

$f'(x)= 8 \cdot\left ( -1 x^{-2} \right )$

$f'(x)= -\frac{8}{x^{ 2}}$

$f'(-2)= -\frac{8}{(-2)^{ 2}} = -\frac{8}{4} = -2$ , the slope of the line.

The equation of the line with slope through (-2,-4) is:

$y - y_{1} = m (x-x_{1})$

y - (-4) = -2 [x- (-2)]

y +4 = -2 (x+2)

y +4 = -2 x-4

y = -2 x-8

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Example Question #4 : Derivative At A Point

What is the equation of the line tangent to the graph of the function

$f(x)= \log (x+3)$

at the point (7,1) ?

Possible Answers:

y = 2x - 13

$y = \frac{x -7 + e \ln 10}{e \ln 10}$

y = x-6

$y = \frac{x + 3 }{10 }$

$y = \frac{x -7 + 10 \ln 10}{10 \ln 10}$

Correct answer:

$y = \frac{x -7 + 10 \ln 10}{10 \ln 10}$

Explanation:

The slope of the line tangent to the graph of $f(x)= \log (x+3)$ at the point (7,1) is f'(7) , which can be evaluated as follows:

$f(x)= \log (x+3)$

$f(x)=\frac{ \ln (x+3) }{\ln 10}$

$f'(x)=\frac{\mathrm{d} }{\mathrm{d} x} \frac{ \ln (x+3) }{\ln 10}$

$f'(x)= \frac{1 }{\ln 10} \cdot \frac{\mathrm{d} }{\mathrm{d} x}\ln (x+3)$

$f'(x)= \frac{1 }{\ln 10} \cdot \frac{1}{x+3}$

$f'(7)= \frac{1 }{\ln 10} \cdot \frac{1}{7+3}$

$f'(7)= \frac{1 }{10 \ln 10}$

The line with this slope through (7,1) has equation:

$y - y_{1} = m (x-x_{1})$

$y - 1= \frac{1 }{10 \ln 10} (x-7)$

$y - 1= \frac{1 }{10 \ln 10} \cdot x- \frac{1 }{10 \ln 10} \cdot 7$ $y = \frac{x }{10 \ln 10} - \frac{7 }{10 \ln 10} +1$

$y = \frac{x -7 + 10 \ln 10}{10 \ln 10}$

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Calculus 2 : Derivatives

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #61 : Derivative Review

Example Question #61 : Definition Of Derivative

Example Question #61 : Derivatives

Example Question #63 : Definition Of Derivative

Example Question #61 : Definition Of Derivative

Example Question #1 : Derivative At A Point

Example Question #2 : Derivative At A Point

Example Question #2 : Derivative At A Point

Example Question #62 : Derivatives

Example Question #4 : Derivative At A Point

All Calculus 2 Resources

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