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AP Calculus BC : Derivative Defined as Limit of Difference Quotient

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Example Question #41 : Derivatives

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

Evaluate g '(6) .

Possible Answers:

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

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

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

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

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

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 #1 : Derivative Defined As Limit Of Difference Quotient

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 =4 x-5

y =2 x-1

y = 3

y = x+1

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 #1 : 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:

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

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

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

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

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

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 #1 : Derivative Defined As Limit Of Difference Quotient

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

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

at (-2,-4) ?

Possible Answers:

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

y = 2 x

y = -2 x-8

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 #61 : Derivative Review

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

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

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

y = x-6

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

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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Example Question #1 : Derivative At A Point

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

$f(x) = x^{3} + x - 17$

at the point (3, 13) ?

Possible Answers:

y = 27 x- 68

y = 11x -20

y = 26 x- 65

y = 10x -17

y = 28 x- 71

Correct answer:

y = 28 x- 71

Explanation:

The slope of the line tangent to the graph of $f(x) = x^{3} + x - 17$ at the point (3, 13) is f'(3) , which can be evaluated as follows:

$f(x) = x^{3} + x - 17$

$f'(x) = \frac{\mathrm{d} }{\mathrm{d} x}\left ( x^{3} + x - 17 \right )$

$f'(x) = 3x^{2} + 1$

$f'(3) = 3\cdot 3^{2} + 1 = 28$

The line with slope 28 through (3, 13) has equation:

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

y -13= 28 (x-3)

$y -13= 28 x- 28 \cdot 3$

y -13= 28 x- 84

y = 28 x- 71

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Example Question #1 : Derivative At A Point

Given the function y=log_e(x)+5x , find the slope of the point (10,51) .

Possible Answers:

$\frac{51}{10}$

The slope cannot be determined.

Correct answer:

$\frac{51}{10}$

Explanation:

To find the slope at a point of a function, take the derivative of the function.

y=log_e(x)+5x

The derivative of log_a(x) is $\frac{1}{x\:ln(a)}$ .

Therefore the derivative becomes,

$\frac{1}{x\:ln(e)}+5=\frac{1}{x}+5$ since ln(e)=1 .

Now we substitute the given point to find the slope at that point.

$\frac{dy}{dx}=\frac{1}{x} +5 = \frac{1}{10}+5 =\frac{51}{10}$

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Example Question #1 : Derivative At A Point

Find the value of the following derivative at the point $\left(2,\frac{1}{2}\right)$ :

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

Possible Answers:

$\frac{1}{4}$

$-\frac{1}{16}$

$\frac{1}{16}$

${}-\frac{1}{4}$

$\frac{1}{\sqrt{\frac{5}{2}}}$

Correct answer:

$-\frac{1}{16}$

Explanation:

To solve this problem, first we need to take the derivative of the function. It will be easier to rewrite the equation as $f(x)=(x+2)^{-\frac{1}{2}}$ from here we can take the derivative and simplify to get

$\\f'(x)=-\frac{1}{2}(x+2)^{-\frac{1}{2}-1}\cdot \frac{\mathrm{d} }{\mathrm{d} x}(x+2)\\ f'(x)=-\frac{1}{2}(x+2)^{-\frac{3}{2}}\cdot 1\\f'(x)=-\frac{1}{2}(x+2)^{-\frac{3}{2}}$

From here we need to evaluate at the given point $\left(2,\frac{1}{2}\right)$ . In this case, only the x value is important, so we evaluate our derivative at x=2 to get $f'(2)=-\frac{1}{2}(2+2)^{-\frac{3}{2}}=-\frac{1}{2}\cdot \frac{1}{8}=-\frac{1}{16}$ .

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Example Question #3 : Derivative At A Point

Evaluate the value of the derivative of the given function at the point (-1,4) :

f(x)=3x^4+2x^2-1

Possible Answers:

Correct answer:

Explanation:

To solve this problem, first we need to take the derivative of the function.

$\\f'(x)=4\cdot 3x^{4-1}+2\cdot 2x^{2-1}\\f'(x)=12x^3+4x$

From here we need to evaluate at the given point (-1,4) . In this case, only the x value is important, so we evaluate our derivative at x=1 to get

$\\f'(-1)=12(-1)^3+4(-1)\\ f'(-1)=12(-1)+4(-1)\\ f'(-1)=-16$ .

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Example Question #7 : Derivative Defined As Limit Of Difference Quotient

Given $f(x)=5x^{2}+6x-12$ , find the value of f'(x) at the point (-1,-1) .

Possible Answers:

Correct answer:

Explanation:

Given the function $f(x)=5x^{2}+6x-12$ , we can use the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$ to find its derivative:

$f'(x)=(2)5x^{2-1}+(1)6x^{1-1}-(0)12=10x+6$ .

Plugging in the -value of the point (-1,-1) into f'(x) , we get

f'(-1)=10(-1)+6=-10+6=-4 .

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AP Calculus BC : Derivative Defined as Limit of Difference Quotient

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #41 : Derivatives

Example Question #1 : Derivative Defined As Limit Of Difference Quotient

Example Question #1 : Derivative At A Point

Example Question #1 : Derivative Defined As Limit Of Difference Quotient

Example Question #61 : Derivative Review

Example Question #1 : Derivative At A Point

Example Question #1 : Derivative At A Point

Example Question #1 : Derivative At A Point

Example Question #3 : Derivative At A Point

Example Question #7 : Derivative Defined As Limit Of Difference Quotient

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