How do you differentiate #f(x)=1/x^2#?

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Answer 1 · 2016-11-03T05:15:15

#d/dxf(x) = -2/ x^3 #

#f(x)= 1/x^2 #
# d/dxf(x) =d/dx(1/x^2) #
# d/dx(1/x^2) = d/dx(x^-2) #
# implies d/dxf(x) = -2* x^-3 #
# implies d/dxf(x) = -2/ x^3 #

Answer 2 · 2016-11-03T05:22:54

#dy/dx=(-2)/x^3#

Given -

#y=1/x^2#

Method -1
The given function can be written as -

#y=x^-2#

#dy/dx=-2x^(-2-1)#

#dy/dx=-2x^-3#

#dy/dx=(-2)/x^3#

Method - 2
We can apply quotient rule

#dy/dx={[(x^2)(0)]-[(1)(2x)]}/((x^2)^2)#

#dy/dx=(-2x)/x^4#

#dy/dx=(-2)/x^3#