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The question is to find the derivative of $ f(x) = (3x^2-3x-6)^\frac{2}{3} $

The answer says :

$ \frac{2}{3}(3x^2-3x-6)^\frac{-1}{3} \cdot (6x-3) $

$ = \frac{2(2x-1)}{(3x^2-3x-6)^\frac{1}{3}} $

How did he get from the first one to the second ?

My answer was :

$ \frac{2(6x-3)}{3(3x^2.3x-6)^\frac{1}{3}} $

Guys i know it's algebra thing so i will appreciate it if you explained what happened here.

And there is another one which was to find the second derivative of $arcsin(2x^3)$

The answers says for the second derivative :

$ \frac{12x \cdot \sqrt{1-4x^6} - 6x^2 \cdot \frac{1}{2}(1-4x^6)^\frac{-1}{2}(-24x^5)}{1-4x^6} $

$ = \frac{12x(1-4x^6)-3x^2(-24x^5)}{(1-4x^6)^\frac{3}{2}} $

So why here the square root disappeared from the first one ?

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$a^{-1} = \dfrac{1}{a}$ and a $3$ factors and cancels from the numerator. Regards – Amzoti Nov 10 '13 at 2:41
Did the answer resolve your issues? – Amzoti Nov 10 '13 at 3:58

We can reduce your answer as:

$$ \dfrac{2(6x-3)}{3(3x^2-3x-6)^\frac{1}{3}} = \dfrac{2 \times 3 (2x-1)}{3(3x^2-3x-6)^\frac{1}{3}} = \dfrac{2(2x-1)}{(3x^2-3x-6)^\frac{1}{3}}$$

A $3$ factors from the numerator and then we can cancel it out.

For the second item, we have:

$ \dfrac{12x \cdot \sqrt{1-4x^6} - 6x^2 \cdot \dfrac{1}{2}(1-4x^6)^\frac{-1}{2}(-24x^5)}{1-4x^6} = \dfrac{12x(1-4x^6)-3x^2(-24x^5)}{(1-4x^6)^\frac{3}{2}} $

What if were to multiply (or, this can be seen as simplifying the numerator) the numerator and denominator by $\sqrt{1-4x^6}$?

Is that clear?

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You're welcome!! +1 – amWhy Nov 10 '13 at 3:08

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