Derivative of a fractional exponent proof?

**cassius** · January 23rd 2011, 06:45 AM

such that if
$f(x) = x^{p/q}$ then $f'(x)={p/q}x^{p/q-1}$

using the definition of a derivative (the 'lim as h tends to zero' one), the formula $a^{n}-b^{n}=(a-b)(a^{n-1}+a^{n-1}b+...+b^{n-1})$
and fact that the derivative of $f(x)=x^{1/q}$ is $1/qx^{1/q-1}$ (proved earlier using the above formula, where the "h" falls out nicely)

I have achieved the result implicitly, but did not use the def-n of the derivative or the $a^{n}-b^{n}$ formula

**Prove It** · January 23rd 2011, 07:17 AM

You can prove that $\displaystyle \frac{d}{dx}(x^n) = n\,x^{n-1}$ for all $\displaystyle n$ using the chain rule.

$\displaystyle y = x^n$

$\displaystyle = e^{\ln{(x^n)}}$

$\displaystyle = e^{n\ln{x}}$ .

Now let $\displaystyle u = n\ln{x}$ so that $\displaystyle y = e^u$ .

$\displaystyle \frac{du}{dx} = \frac{n}{x}$ .

$\displaystyle \frac{dy}{du} = e^u = e^{n\ln{x}}$ .

Thus $\displaystyle \frac{dy}{dx} = \frac{du}{dx}\,\frac{dy}{du}$

$\displaystyle = \frac{n\,e^{n\ln{x}}}{x}$

$\displaystyle = \frac{n\,e^{\ln{(x^n)}}}{x}$

$\displaystyle = \frac{n\,x^n}{x}$

$\displaystyle = n\,x^{n-1}$ .

Q.E.D.

Of course, you will need to have proved the chain rule and the derivatives of $\displaystyle e^x$ and $\displaystyle \ln{x}$ beforehand...

**cassius** · January 23rd 2011, 07:32 AM

thank you for this, but i am required to construct a proof from first principles and only using the information given

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