Question

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

Answer 1

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)`

Answer 2

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`