How do you find the derivative of #1/x^2#?

2 Answers
Mar 30, 2015

let #y = 1/(x^2)#
#1/(x^2)# is the same as #x^(-2)#
#=> (dy)/(dx) = -2*x^(-2-1)#
#= -2x^(-3)#
#=> (dy)/(dx) = -2/(x^-3)#

Mar 30, 2015

To find the derivative of a ratio, apply the formula
#D f(x)/g(x)= {f'(x)g(x)-f(x) g'(x)}/{g^2(x)}#
In particular, if #f(x)=1#, you have that #f'(x)=0#, and the expression becomes
#D 1/g(x) = {-g'(x)}/{g^2(x)}#

Since your #g(x)# is #x^2#, you easily get that #g'(x)=2x#. So,
#{-g'(x)}/{g^2(x)}=-{2x}/{x^4}=-2/x^3#