Question

Find f'(x) of f(x)=1/x^2

Answer 1

know the derivative of x^n ?

Answer 2

Yes

Answer 3

then 1/x^2 is just x^-2
plug in n=-2 in that formula

Answer 4

Yeah I know that but I'm suppose to use the formula f'(x)=h-> 0f(x+h)-f(x)/h

Answer 5

ok, then whats f(x+h) = .. ?

Answer 6

1/(x+h)^2

Answer 7

yes,
so, your numerator
f(x+h) - f(x)
will be
1/ (x+h)^2 + 1/x^2
try to simplify this ? by making a common denominator?

Answer 8

soory, there should be a minus - in between

Answer 9

1/ (x+h)^2 - 1/x^2

Answer 10

umm would it be x^2/(x+h)^2x^2 - (x+h)^2/x^6(x+h)^2/h

Answer 11

how is it x^6 there ?
did you mean x^2 ?
everything else seem to be correct

Answer 12

so we have a common denominator of x^2 (x+h)^2 now , right ?
when we combine the numerator, we get
\(\Large \dfrac{x^2-(x+h)^2}{h x^2(x+h)^2}\)
did you get this ?

Answer 13

oh sorry x^2

Answer 14

yes i did

Answer 15

simplify the numerator, anything getting cancelled ?

Answer 16

umm the x^2?

Answer 17

correct!
from whatever remains,
factor out the "h"

Answer 18

so h(2x+h)/hx^2(x+h)^2?

Answer 19

there will be a minus sign, right ?
\(\Large \dfrac{x^2-x^2 -2xh-h^2}{h x^2(x+h)^2}= \Large \dfrac{-\cancel h (2x+h)}{\cancel h x^2(x+h)^2}\)
got this ?

Answer 20

got how that "h" got cancelled ?

Answer 21

ohh yesss i see lol i forgot the minus sign

Answer 22

so what happens to the h in the denominator?

Answer 23

one of the 'h' got cancelled, right
now since we have limit as h->0
we can directly plug in h =0 in the function now! :)
what do u get ?

Answer 24

2x?

Answer 25

numerator = -2x yes,
what about the denominator, when h= 0 ?

Answer 26

hmm im not sure..

Answer 27

the denominator is x^2 (x+h)^2
right ?
so what u get when u put h= 0 ?

Answer 28

x^4?

Answer 29

yes!
so we have
-2x/x^4
what does that simplify to ?

Answer 30

-2x^-3?

Answer 31

absolutely correct!
derivative of 1/x^2 is indeed -2/x^3 or -2x^-3

Answer 32

yay! thank you so so much you made this 10 times easier to understand ^-^

Answer 33

i am glad to hear that :)
you're most welcome ^_^