Question

Howcome 1/x^2 equals -2/x^3? According to the calculator I use, it utilizes the power rule...-But how does that work when we have 1/something and not x^something? Thanks ahead for taking the time to assist :)

Answer 1

Please see below. (Short answer: because 1/x^something = x^minus something.)

Explanation:

First thing: `1/x^2` is NOT equal to `(-2)/x^3`.

The derivative of `1/x^2` is equal to `(-2)/x^3`.

Second:

We can write `y = 1/x^2` as `y = x^(-2)`.

When we apply the power rule we get

`y' = -2x^(-2-1) = -2x^(-3)`

Now we can rewrite that as

`= -2 * 1/x^3 = (-2)/x^3`

Bonus

Can you see why the derivative of `y = 1/x^5` is

`y' = (-5)/x^6`?

Answer 2

By the rule of indices:

`1/x^n =x^(-n)`

We know that for positive integers, `n in NN` that:

`d/dx x^n =nx^(n-1)`

Now for the case where `n` is a negative integer, put `m=-n` so that `m in NN`, then we have:

`d/dx 1/x^n = d/dx x^(-n)`
`\ \ \ \ \ \ \ \ \ \ \ = d/dx x^(m)`
`\ \ \ \ \ \ \ \ \ \ \ = mx^(m-1)` (as `m in NN`)
`\ \ \ \ \ \ \ \ \ \ \ = -nx^(-n-1)`

Proving that

`d/dx x^n =nx^(n-1) AA n in ZZ^+`

Which then gives us the result with `n=-2`, that:

`d/dx 1/x^2 = (-2)x^(-3) = -2/x^3` QED