Diferentiation

**stripe501** · March 1st 2011, 04:06 AM

How do you differentiate function like x^(1/4) or x^(3/5) ect?

**stripe501** · March 1st 2011, 04:15 AM

never mind lololol, d/dx=nx^(n-1)

for
f(x)=x^(1/4)
f'(x)=(1/4)*x^(-3/4)

**chisigma** · March 1st 2011, 06:17 AM

Originally Posted by stripe501

How do you differentiate function like x^(1/4) or x^(3/5) ect?

Given $f(x)= x^{\alpha}$ , where $\alpha$ is an arbitrary real or even complex constant, its derivative by definition is...

$\displaystyle f^{'}(x) = \lim_{\delta \rightarrow 0} \frac{(x+\delta)^{\alpha} - x^{\alpha}}{\delta}$ (1)

Now is...

$\displaystyle {(x+\delta)^{\alpha} = x^{\alpha} + \alpha\ \delta\ x^{\alpha-1} + \frac{\alpha\ (\alpha-1)}{2}\ \delta^{2}\ x^{\alpha-2} + ...$ (2)

... so that, combining (1) and (2), we find that is...

$\displaystyle f^{'}(x)= \alpha\ x^{\alpha-1}$ (3)

Kind regards

$\chi$ $\sigma$

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