# Thread: Differentiating with roots and powers.

1. ## Differentiating with roots and powers.

Hi all,

I have started reading Calculus Made Easy by Silvanus P.Thompson and am having a little difficulty early in.

The problem is

$y=\sqrt[q]{x^{3}}$

and I have $\frac{dy}{dx}=\frac{3}{q}x^{\frac{3}{q}-1}$

but the solution is $\frac{3}{q}x^{\frac{3-q}{q}}$

My workings are

$y=\sqrt[q]{x^{3}}=x^{\frac{3}{q}}, \frac{dy}{dx}=\frac{3}{q}x^{\frac{3}{q}-1}$

I thought that during the differentiation the index is reduced by 1?

Any help would be great,

thanks.

2. ## Re: Differentiating with roots and powers.

Originally Posted by Srengam
Hi all,

I have started reading Calculus Made Easy by Silvanus P.Thompson and am having a little difficulty early in.

The problem is

$y=\sqrt[q]{x^{3}}$

and I have $\frac{dy}{dx}=\frac{3}{q}x^{\frac{3}{q}-1}$

but the solution is $\frac{3}{q}x^{\frac{3-q}{q}}$

My workings are

$y=\sqrt[q]{x^{3}}=x^{\frac{3}{q}}, \frac{dy}{dx}=\frac{3}{q}x^{\frac{3}{q}-1}$

I thought that during the differentiation the index is reduced by 1?

Any help would be great,

thanks.
The two answers are the same (and correct). The book have wrote 1 as q/q.

$\dfrac{3}{q} -1 = \dfrac{3}{q} - \dfrac{q}{q} = \dfrac{3-q}{q}$

3. ## Re: Differentiating with roots and powers.

Many thanks, that helps a lot.