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Mathematics 16 Online
OpenStudy (anonymous):

Find the derivative of f(x) = 8 divided by x at x = -1.

OpenStudy (anonymous):

@ganeshie8 @robtobey if you could assit me i would appreciate your help.

ganeshie8 (ganeshie8):

familiar with any formulas ?

OpenStudy (anonymous):

would you have to do the difference quotient formula

ganeshie8 (ganeshie8):

yes, can you show me the formula which you have ?

OpenStudy (anonymous):

f(x+h)-f(x)/h

OpenStudy (anonymous):

this is where i dont know how to proceed

ganeshie8 (ganeshie8):

what about the limit ?

ganeshie8 (ganeshie8):

\[\large f'(x) = \lim \limits_{h\to 0} \dfrac{f(x+h) - f(x)}{h}\]

OpenStudy (anonymous):

the limit is -1 in the formula its limit h to 0=f(x+h)-f(x)/h

ganeshie8 (ganeshie8):

thats the formula right ?

OpenStudy (anonymous):

yes exactly

ganeshie8 (ganeshie8):

start by finding the difference `f(x+h) - f(x)`

ganeshie8 (ganeshie8):

f(x) = 8/x f(x+h) = ?

OpenStudy (anonymous):

8(x+h) -8/x

ganeshie8 (ganeshie8):

yes, put them as single fraction by working the common denominator

OpenStudy (anonymous):

x(8(x+h))-8/x

OpenStudy (anonymous):

8x(x^2+hx)-8/x

ganeshie8 (ganeshie8):

\(\large \dfrac{8}{x+h} - \dfrac{8}{x}\)

ganeshie8 (ganeshie8):

multiply the first fraction by x/x second fraction by (x+h)/(x+h)

ganeshie8 (ganeshie8):

\(\large \dfrac{x}{x}\dfrac{8}{x+h} - \dfrac{8}{x}\dfrac{x+h}{x+h}\)

ganeshie8 (ganeshie8):

\(\large \dfrac{8x}{x(x+h)} - \dfrac{8(x+h)}{x(x+h)}\)

ganeshie8 (ganeshie8):

Now that the denominators are same, you can add/subtract the numerators

ganeshie8 (ganeshie8):

\(\large \dfrac{8x - 8(x+h)}{x(x+h)}\)

ganeshie8 (ganeshie8):

simplify

OpenStudy (anonymous):

8h/x^2+hx

ganeshie8 (ganeshie8):

careful about signs *

OpenStudy (anonymous):

-8h/x^2+hx

ganeshie8 (ganeshie8):

yes, so f(x+h) - f(x) = -8h/(x^2+hx)

ganeshie8 (ganeshie8):

plug this value in your earlier difference quotient formula

ganeshie8 (ganeshie8):

\[\large f'(x) = \lim \limits_{h\to 0} \dfrac{f(x+h) - f(x)}{h}\]

ganeshie8 (ganeshie8):

\[\large f'(x) = \lim \limits_{h\to 0} \dfrac{-8h/(x^2+hx)}{h}\]

ganeshie8 (ganeshie8):

h cancels out ^

ganeshie8 (ganeshie8):

\[\large f'(x) = \lim \limits_{h\to 0} -8/(x^2+hx)\]

ganeshie8 (ganeshie8):

Since there is no h in the bottom, you can take the limit now

ganeshie8 (ganeshie8):

simply plugin h = 0

ganeshie8 (ganeshie8):

\[\large f'(x) = \lim \limits_{h\to 0} -8/(x^2+hx)\] \[\large f'(x) = -8/(x^2+0 \times x)\] \[\large f'(x) = -8/x^2\]

ganeshie8 (ganeshie8):

thats the derivative function ^

ganeshie8 (ganeshie8):

plugin x = -1 to get the derivative at x = -1

OpenStudy (anonymous):

would it -8 as your derivative

ganeshie8 (ganeshie8):

Yep !

OpenStudy (anonymous):

thank you for your help its very kind of you to take the the time and effort to assist me. i appreciate your assistance and i hope you have a nice day.

ganeshie8 (ganeshie8):

np, you're welcome :) good day !

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