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

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

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

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

My workings are

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

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

Any help would be great,

thanks.

Re: Differentiating with roots and powers.

Quote:

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

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

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

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

My workings are

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

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.

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

Re: Differentiating with roots and powers.

Many thanks, that helps a lot.