derivatives of trigo identity

**jasonlewiz** · August 22nd 2010, 05:52 AM

I don't how to start with this kind of problem. Can someone show me the solution for this one.

Find the first derivative

sin^2(sin(sin x))

**Archie Meade** · August 22nd 2010, 06:09 AM

Originally Posted by jasonlewiz

I don't how to start with this kind of problem. Can someone show me the solution for this one.

Find the first derivative

sin^2(sin(sin x))

$u=sinx$

$v=sinu$

$w=sin^2v$

$\displaystyle\frac{dw}{dx}=\frac{dw}{dv}\ \frac{dv}{du}\ \frac{du}{dx}$

$=\displaystyle\frac{d}{dv}sin^2v\ \frac{d}{du}sinu\ \frac{d}{dx}sinx$

**Archie Meade** · August 22nd 2010, 03:45 PM

If you like you can write....

$u=Sinx$

$v=Sinu$

$w=Sinv$

$t=w^2$

then, using the Chain Rule of differentiation...

$\displaystyle\frac{dt}{dx}=\frac{dt}{dw}\ \frac{dw}{dv}\ \frac{dv}{du}\ \frac{du}{dx}$

$=\displaystyle\frac{d}{dw}w^2\ \frac{d}{dv}Sinv\ \frac{d}{du}Sinu\ \frac{d}{dx}Sinx$

$=[2w][Cosv][Cosu][Cosx]=[2Sinv][Cos(Sinu)][Cos(Sinx)][Cosx]$

$=[2Sin(Sinu)][Cos(Sin(Sinx))][Cos(Sinx)][Cosx]$

$=[2Sin(Sin(Sinx))][Cos(Sin(Sinx))][Cos(Sinx)][Cosx]$

**tom@ballooncalculus** · August 24th 2010, 02:15 AM

Just in case a picture helps...

... where (key in spoiler) ...

Spoiler:

Alternatively...

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