# Thread: derivatives of trigo identity

1. ## 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))

2. 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))
$\displaystyle u=sinx$

$\displaystyle v=sinu$

$\displaystyle w=sin^2v$

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

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

3. If you like you can write....

$\displaystyle u=Sinx$

$\displaystyle v=Sinu$

$\displaystyle w=Sinv$

$\displaystyle t=w^2$

then, using the Chain Rule of differentiation...

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

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

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

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

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

4. Just in case a picture helps...

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

Spoiler:

... 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).

Spoiler:

Alternatively...

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Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

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