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

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

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

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

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