AP Calculus BC - Derivative Defined as Limit of Difference Quotient

Find the derivative of at point .

Explanation

Use either the FOIL method to simplify before taking the derivative or use the product rule to find the derivative of the function.

The product rule will be used for simplicity.

Substitute .

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

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 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 into , we get

If $f(x)=\ln(2x)$ , which of the following limits equals ?

$\lim_{h \to 0} \frac{\ln(e+1+h)-\ln(e+1)}{h}$

$\lim_{h \to 0} \frac{\ln(e+1+h)-\ln(e+1)}{e+1}$

$\lim_{h \to \infty} \frac{\ln(e+1+h)-\ln(e+1)}{e+1}$

$\lim_{h \to 0} \frac{\ln(1+h)-\ln(h)}{h}$

$\lim_{h \to \infty} \frac{\ln(e+1+h)-\ln(e+1)}{h}$

Explanation

The equation for the derivative at a point is given by

$f'(a)=\lim_{h \to 0}\frac{f(a+h)-f(a)}{h}$ .

By substituting , $f = \ln(2x)$ , we obtain

$f'(e+1)=\lim_{h \to 0} \frac{\ln(e+1+h)-\ln(e+1)}{h}.$

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

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

at the point ?

Explanation

The slope of the line tangent to the graph of $f(x) = x^{3} + x - 17$ at the point is , 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 has equation:

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

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

What is the slope of the tangent line to the function

$h(x)=e^{x}ln(x)$

when

$\frac{1}{e}$

Explanation

The slope of the tangent line to a function at a point is the value of the derivative at that point. To calculate the derivative in this problem, the product rule is necessary. Recall that the product rule states that:

$\frac{d}{dx}(m(x)n(x))=m'(x)n(x)+m(x)n'(x)$ .

In this example, $m(x)=e^x \ and\ n(x)=ln(x).$

Therefore,

$m'(x)=e^x \ and\ n'(x)=\frac{1}{x}$ , and

$\frac{d}{dx}(e^xln(x))=e^xln(x)+e^x\cdot\frac{1}{x}$

At x = 1, this dervative has the value

$e^1ln(1)+e^1\cdot\frac{1}{1}=0+e=e$ .

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 ?

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

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

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

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

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

Explanation

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

, 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 is:

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

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

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

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

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

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

at the point ?

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

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

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

Explanation

The slope of the line tangent to the graph of $f(x)= \log (x+3)$ at the point is , 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 has equation:

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

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

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

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

Find the derivative of the following function at :

$h(x)=2x\sin(x)$

$2\sin(2)+4\cos(2)$

$2\sin(2)-4\cos(2)$

$2\sin(2)$

$2\cos(2)$

Explanation

The derivative of the function is given by the product rule:

Simply find the derivative of each function:

$\frac{d}{dx}(2x)=2$

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

The derivatives were found using the following rules:

$\frac{d}{dx}\sin(x)=\cos(x)$ , $\frac{d}{dx}(x^n)=nx^{n-1}$

Simply evaluate each derivative and the original functions at the point given, using the above product rule.

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

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

at ?

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

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

Explanation

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

, 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 is:

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

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

Evaluate .

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

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

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

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

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

Explanation

To find , 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

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

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

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

Derivative Defined as Limit of Difference Quotient

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