Question

If `y = 8/(3x^2)`, what is the value of `y'`?

Answer 1

There is no problem that I can see in the reasoning used.

If you check your answer using the quotient rule, you get the same derivative.

Here's the check:

`y' = (0 xx 3x^2 - 6x xx 8)/(3x^2)^2`

`y' = (-48x)/(9x^4)`

`y' = -16/(3x^3)`

We get the same derivative (since `-16/3x^(-3) = -16/(3x^3))`, using a different method.

Hopefully this helps!

Answer 2

There are many routes to the correct answer. Your method is fine.

Explanation:

I would have thought of the details this way:

`8/(3x^2) = 8/3 * 1/x^2 = 8/3x^-2`.

I get the exact from you got:

`y' = -16/3x^-3`.

Another possibility (that I find needlessly complicated) is to use the product rule on `8/3*x^-2`

`y' = d/dx(8/3) * x^-2 + 8/3 * d/dx(x^-2)`

`= 0 * x^-2 + 8/3 * (-2x*-3)`

`= -16/3x^-3`.

Needlessly complicated? I think so.

Incorrect? No it is not incorrect.

(If I try to go to Chicago, there are many ways to get there. If I end up in Chicago, the way I chose worked.)