# Math Help - Differentiation of Algebraic Function

1. ## Differentiation of Algebraic Function

Hello all.
I'm starting calculus and I'm stumped by this exercise. I've arrived at multiple answers only to realise that, according to my text, I'm lost.

Find the derivative of:
$\frac{x^3}{\sqrt[3]{3x^2-1}}$

Here is my working:

Change fractions to exponents.
$x^3(3x^2-1)^\frac{-1}{3}$

Apply the product rule and then the chain rule for the derivative of the bracketed (3x...) term.
$3x^2(3x^2-1)^\frac{-1}{3} + x^3(-2x(3x^2-1)^\frac{-4}{3})$

Expand.
$\frac{3x^2}{(3x^2-1)^\frac{1}{3}} - \frac{2x^4}{(3x^2-1)^\frac{4}{3}}$

At that point it just doesn't seem right. I'm way off track according to the book's answer but I'm just not sure where I'm going wrong.

Any help would be much appreciated.

2. Originally Posted by jerry
Hello all.
I'm starting calculus and I'm stumped by this exercise. I've arrived at multiple answers only to realise that, according to my text, I'm lost.
Find the derivative of:
$\frac{x^3}{\sqrt[3]{3x^2-1}}$
Here is my working:

Change fractions to exponents.
$x^3(3x^2-1)^\frac{-1}{3}$

Apply the product rule and then the chain rule for the derivative of the bracketed (3x...) term.
$3x^2(3x^2-1)^\frac{-1}{3} + x^3(-2x(3x^2-1)^\frac{-4}{3})$

Expand.
$\frac{3x^2}{(3x^2-1)^\frac{1}{3}} - \frac{2x^4}{(3x^2-1)^\frac{4}{3}}$

At that point it just doesn't seem right. I'm way off track according to the book's answer but I'm just not sure where I'm going wrong.

Any help would be much appreciated.
Find an LCD. In this case, its $(3x^2-1)^{\frac{4}{3}}$

So $\frac{3x^2}{(3x^2-1)^\frac{1}{3}} - \frac{2x^4}{(3x^2-1)^\frac{4}{3}}\implies \frac{3x^2}{(3x^2-1)^\frac{1}{3}}\cdot{\color{red}\frac{3x^2-1}{3x^2-1}} - \frac{2x^4}{(3x^2-1)^\frac{4}{3}}$ $\implies\frac{3x^2(3x^2-1)}{(3x^2-1)^\frac{4}{3}} - \frac{2x^4}{(3x^2-1)^\frac{4}{3}}$ $\implies\frac{9x^4-3x^2-2x^4}{(3x^2-1)^\frac{4}{3}}\implies\color{red}\boxed{\frac{7x^ 4-3x^2}{(3x^2-1)^\frac{4}{3}}}$

Do they have this as the answer?

--Chris

3. Same result Chris showed but using logaritmic differentiation, it can be quite useful in some cases:

y = x^3/[(3x^2-1)^(1/3)]

Ln y = Ln|x^3| - Ln|(3x^2-1)^(1/3)|

(1/y)(dy/dx) = 3/x - 2x/(3x^2-1)

(1/y)(dy/dx) = (7x^2-3)/(x(3x^2-1))

(dy/dx) = (7x^4-3x^2) / (3x^2-1)^(4/3)

4. Thank you Chris! That's the answer. So it was a small oversight all along. And thanks for the interesting alternative dax. I'll be sure to take a look at it.