# HEllo!!! I know its late but does anyone know how to solve derivatives? ?

im so lame I barely got a C on precalculus and know im taking Calculus

okay so this is the problem

if f(x)= 3/x^2, find f ' (4) using definition of derivative.

f ' (4) is the limit as

x --> ___ (this where i to put something) of the expression ___.

the value of the limit is _____ .

the best guess i can make is to use [ f(x+h) - f(x) ] / h and the substitute 4 somewhere but i dont know is it h or x??

if any one can show the steps to how to solve this id be so grateful

Relevance
• Anonymous

Definition of Derivative is

Lim [ f(x+h) - f(x) ] / h

h--->0

Lim [ 3/(x+h)^2 - 3/x^2 ] / h

h--->0

Lim [ (3x^2-3(x+h)^2)/(x+h)^2x^2 ] / h

h--->0

Lim [ (3x^2-3(x+h)^2)/(x+h)^2x^2 ] / h

h--->0

Lim [ 3x^2-3x^2 -6xh -3h^2]/(x+h)^2x^2 h

h--->0

Lim [ -6xh -3h^2)]/(x+h)^2x^2 h

h--->0

Lim [ h(-6x -3h)]/(x+h)^2x^2 h

h--->0

Lim [ (-6x -3h)]/(x+h)^2x^2

h--->0

Plug in h =0

-6x/x^4 = -6/x^3

Plug in x=4

f ' (4) = -6/4^3 = -6/64 = -3/32

Also to get the ans quickly

If f(x) = x^n

f ' (x) = nx^n-1

f ' (x ) = -6/x^3 , f ' (4 ) = -6/4^3 = -6/64 = -3/32