how do u find f'(x): f(x) = 1/x^2?

Answer 1

Now you can use the power rule which has already been demonstrated by a lot of people, or you can use "quotient Rule" which might seem a little difficult but in the long run it will help you more then power rule in this kinds of questions (a/b) because it uses power rules aswell. So Mary the rule is as following:

f (u / v) '= [ u' (v) - u (v')] / [(v)^2]
f ' = [ (0)(x^2) - (1)(2x)] / [(x^2)2]
f ' = -2x / x^4
f ' = -2x^(1-4)
f ' = -2x^(-3)
f ' = (-2) / (x^3)
hope this helps:)

Answer 2

if f(x) = x^n , then f'(x)= n*x^(n-1)
f(x) = 1/x^2 = x^(-2)
f'(x) = -2* x^(-2-1)
= -2*x^(-3)
= -2/x^3

Answer 3

f (x) = x^( - 2 )
f `(x) = ( - 2 ) x^(-3)
f `(x) = ( - 2 ) / x ³

Answer 4

For differentiating any variable with power n, the general formula is n*{x^(n-1)}...
So for the given problem since you have the variable in the denominator, you have to take it as x^(-2)..
And apply the formula..
You will have -2*x^(-2-1)= -2x^(-3)

Answer 5

f(x) = 1/x^2
= x^(-2)
then
f'(x) = -2 x^(-3)*d/dx (x)
= -2 x^(-3)*1
=-2/x^3

Answer 6

the standard formula for differentiating f(x)=X^n is f'(x) =nx^(n-1)
For given problem
f(x) =x^-2 implying
f'(x) =-2x^-3

Answer 7

f(x)=1/x^2 ie x^-2
therefore f'(x)=-2{x^(-2-1)}
ie f'(x)=-2/x^3

Answer 8

f'(x) = (x^-2)' = -2x^-3 = -2/x^3