Quote:

Originally Posted by **Bert**

Hello,

I have here a strength differential $\displaystyle \frac{-t}{\sqrt{1-t^2}}$

Normally is not a big problem but I making a mistake I don't know:

let my try $\displaystyle u=1-t^2$ thus $\displaystyle u'=-2t$

$\displaystyle v=\sqrt{u}$ thus $\displaystyle v'=\frac{1}{2}u^{-\frac{1}{2}$

And $\displaystyle w=\frac{-t}{v}$ thus $\displaystyle w'=\frac{-v+tv'}{v^2}$

Now I need to multiply al the thinks and then I find $\displaystyle \frac{t}{1-t^2}$ but it's not correct.

Where do I miss ? Greets.

Stage 1 - Product rule:

$\displaystyle

\frac{d}{dt}\left(\frac{-t}{\sqrt{1-t^2}}\right)=\frac{1}{\sqrt{1-t^2}}\frac{d}{dt} (-t)+(-t)\frac{d}{dt} \ \left(\frac{1}{\sqrt{1-t^2}}\right)

$

$\displaystyle

=-\frac{1}{\sqrt{1-t^2}}-t\ \frac{d}{dt} \ \left(\frac{1}{\sqrt{1-t^2}}\right)

$

Stage 2 - Chain rule on last term on the right:

$\displaystyle

=-\frac{1}{\sqrt{1-t^2}}-t\ (-1/2)\frac{1}{(1-t^2)^{3/2}}\frac{d}{dt}(1-t^2)

$

$\displaystyle

=\frac{1}{\sqrt{1-t^2}}-t\ (-1/2)\frac{1}{(1-t^2)^{3/2}}(-2t)

$

$\displaystyle

=\frac{1}{\sqrt{1-t^2}}-\frac{t^2}{(1-t^2)^{3/2}}

$

From here you should be able to simplify this yourself.

RonL