How to find a 1st derivative? ?

Question

Hello

I have a question where I have to "find the first derivative of the following function, simplifying where possible". Could anyone explain in the simplest of terms how to do this?

y=³√ 3x²-9x+6

I have an already worked answer but I am unsure how to follow it:

y=(3x^2-9x+6)^(1/3) <-- I assume because ³√ is basically 1/3 power?

dy/dx = (1/3) (3x^2-9x+6)^(1/3-1) times (6x-9) <--- So why is there now one (1/3), and one (1/3-1)?

dy/dx = (6x-9)/ 3(3x^2-9x+6)^(2/3) <--- So how is it now 2/3? What happened to the first (1/3)? Why do we now have the (6x-9) OVER the rest of it?

dy/dx = 3(2x-3) / 3 (3x^2-9x+6)^(2/3) <--- Why is the (6x-9) now 3(2x-3)?

dy/dx = (2x-3) / (3x^2-9x+6) <--- Have both 3s now cancelled out now? Is that then the final answer?

Thanks in advance!! (I'm rubbish at math)

Answer 1

the answer is (2x-3) / (3x^2 -9x +6 )^(2/3).

OK...Yes the cube root is the 1/3 power.

When you take the deriv. of [f(x)]^n he answer is n [f(x)]^(n-1) * f&#039;(x)

so the first 1/3 is the cube root 1/3 swinging out to the front, and

the (1/3-1)=2/3 is the &quot;n-1&quot; power that remains.  Since it was a cube root, it&#039;s 1/3 munus 1=2/3  See?  The first 1/3 remains there, it&#039;s just that they put the 3 in the denominator.  E.g., (1/3) x = x/3 See?

And the 6x-9 is multiplied at the end, so it can be written in front in the numerator.  E.g.  (x-7)/3 * (6x-9) = (6x-9)(x-7)/3  See?

And they factor the 6x-9=3(2x-3) so that this 3 and the 3 in the denominator from the 1/3 previously would cancel, eliminating both

the 3 in the numerator and the 3 in the denom.

ANSWER