1. ## Differential

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.

2. 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

3. thus you can splits this in two parts by the produc rule

then you have $\displaystyle \frac{1}{\sqrt{1-t^2}}$

I can find then on this way: $\displaystyle u=1-t^2$ Thus $\displaystyle u'=-2t$

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

Then $\displaystyle \frac{\frac{-t}{\sqrt{1-t^2}}}{\sqrt{1-t^2}^2}=\frac{-t}{\sqrt{1-t^2}}\frac{1}{1-t^2}$

But why is this fault ? $\displaystyle u=1-t^2 \rightarrow u'=-2t \ \ \ v=\sqrt{u} \rightarrow\frac{1}{2}u^{-\frac{1}{2}} \ \ \ w=\frac{1}{v} \rightarrow w'=lnv$

And then multiply them what is mis? Greets.

4. Originally Posted by Bert
But why is this fault ? $\displaystyle u=1-t^2 \rightarrow u'=-2t \ \ \ v=\sqrt{u} \rightarrow\frac{1}{2}u^{-\frac{1}{2}} \ \ \ w=\frac{1}{v} \rightarrow w'=lnv$

And then multiply them what is mis? Greets.
Mistake that if,
$\displaystyle w=\frac{1}{v}$
then,
$\displaystyle w'=-\frac{1}{v^2}$
You confused with anti-derivative.

5. of cours thank you very well.

a small problem If I simplify $\displaystyle \frac{-t}{\sqrt{1-t^2}}\frac{1}{1-t^2}$

then I get $\displaystyle -\frac{t}{{(1-t^2)}^{\frac{3}{2}}}}$ And this it not the same as I differentiate the original by my computer program then I get $\displaystyle -\frac{1}{{(1-t^2)}^\frac{3}{2}}$

6. Originally Posted by Bert
thus you can splits this in two parts by the produc rule
then you have $\displaystyle \frac{1}{\sqrt{1-t^2}}$
I can find then on this way: $\displaystyle u=1-t^2$ Thus $\displaystyle u'=-2t$

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

...

Hello,

Then $\displaystyle v=\sqrt{u}$ and $\displaystyle v'=\frac{1}{2}u^{-\frac{1}{2}}\ \gets \mbox{this looks funny to me}$

If you use the chain rule on v' then you get: $\displaystyle v'=\frac{1}{2}u^{-\frac{1}{2}} \cdot (-2t)=-t \cdot u^{-\frac{1}{2}}$

Greetings

EB

7. Thank you very well I find them great good forum Greets Thanks