Directions: Find the derivaive of f(x) at x. f(x)= 1/(x+2) x= -1 I used the formula f^1(x)= Limit as h approaches 0 [f(x+h) - f(x)]/h And now I'm at [ 1/(x+h+2) - 1/(x+2) ] /h I would appreciate any help.
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Originally Posted by Purpledog100 Directions: Find the derivaive of f(x) at x. f(x)= 1/(x+2) x= -1 I used the formula f^1(x)= Limit as h approaches 0 [f(x+h) - f(x)]/h And now I'm at [ 1/(x+h+2) - 1/(x+2) ] /h I would appreciate any help. From there, combine the fractions to get $\displaystyle \lim_{h\to0}\frac{x+2-(x+h+2)}{h(x+2)(x+h+2)}=\lim_{h\to0}\frac{-h}{h(x+2)(x+h+2)}=\dots$ Can you continue?
Do the instructions specify that you need to use the limit definition of the derivative? If not, just use the power rule. $\displaystyle \frac{dy}{dx}=-\frac{1}{(x+2)^{2}}$ $\displaystyle \frac{dy}{dx}(-1)=-\frac{1}{(-1+2)^{2}}=-\frac{1}{(1)^{2}}=-1$
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