# derivatives of trigo identity

• Aug 22nd 2010, 06:52 AM
jasonlewiz
derivatives of trigo identity
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))
• Aug 22nd 2010, 07:09 AM
Quote:

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))

$u=sinx$

$v=sinu$

$w=sin^2v$

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

$=\displaystyle\frac{d}{dv}sin^2v\ \frac{d}{du}sinu\ \frac{d}{dx}sinx$
• Aug 22nd 2010, 04:45 PM
If you like you can write....

$u=Sinx$

$v=Sinu$

$w=Sinv$

$t=w^2$

then, using the Chain Rule of differentiation...

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

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

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

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

$=[2Sin(Sin(Sinx))][Cos(Sin(Sinx))][Cos(Sinx)][Cosx]$
• Aug 24th 2010, 03:15 AM
tom@ballooncalculus
Just in case a picture helps...

http://www.ballooncalculus.org/draw/diffChain/nest.png

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

Spoiler:
http://www.ballooncalculus.org/asy/chain.png

... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case, initially x, then u, then later v), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

Alternatively...

http://www.ballooncalculus.org/draw/diffChain/nestb.png
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