Hi

$\displaystyle t^{{\alpha}-1} dt$

When you differentiate that, its not

$\displaystyle ({\alpha}-1)t^{{\alpha}-2$ ? Surely it is just $\displaystyle 0$ ?

Or am I just being completely stupid..

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- Mar 14th 2012, 08:18 AMramdropAm I being really stupid or what? (Differentiation)
Hi

$\displaystyle t^{{\alpha}-1} dt$

When you differentiate that, its not

$\displaystyle ({\alpha}-1)t^{{\alpha}-2$ ? Surely it is just $\displaystyle 0$ ?

Or am I just being completely stupid.. - Mar 14th 2012, 09:03 AMtom@ballooncalculusRe: Am I being really stupid or what? (Differentiation)
$\displaystyle \frac{\mathrm{d}}{\mathrm{d}t}\ t^{a}\ =\ at^{a-1}$

(where t is the independent variable and a is a constant power)

or as we usually say,

$\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}\ x^{n}\ =\ nx^{n-1}$

I haven't twigged why you want it to be 0. - Mar 14th 2012, 09:06 AMramdropRe: Am I being really stupid or what? (Differentiation)
Nvm Solved it, its beena long day and I just made a stupid error xD