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February 27th, 2017, 10:13 AM   #1
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Chain Rule?

I'm trying to learn this stuff by myself. Can someone explain to me how to find the derivative of "sin(sin(sin(sin(sin(x)))))" using the Chain Rule? I'm confused by the whole concept.

Thank you.
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February 27th, 2017, 10:33 AM   #2
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how about just $\sin(\sin(x))$

that should give you the idea

let's work it out generally first though, suppose we have

$\dfrac{d}{dx}~f(g(x))$

first we just find $\dfrac{df}{dx}$ just treating $g(x)$ as a constant

Then we have to multiply this by $\dfrac{dg}{dx}$ and we end up with

$\dfrac{d}{dx}~f(g(x)) = \dfrac{df}{dx}(g(x)) \cdot \dfrac{dg}{dx}(x)$

to really drive things home consider

$\dfrac{d}{dx}~f(g(h(x))) = \dfrac{df}{dx}(g(h(x)))\cdot \dfrac{dg}{dx}(h(x)) \cdot \dfrac{dh}{dx}(x)$

now to your abbreviated example.

$\dfrac{d}{dx}~\sin(\sin(x)) = \cos(\sin(x))\cdot \cos(x)$

eh we can go one further

$\dfrac{d}{dx}~\sin(\sin(\sin(x))) = \cos(\sin(\sin(x)))\cdot \cos(\sin(x)) \cdot \cos(x)$

see the pattern?
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March 1st, 2017, 01:21 AM   #3
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According to @rosmek solution next steps will be-

$=\cos(\sin(\sin(\sin(x))))\cdot \cos(\sin(\sin(x)))\cdot\cos( \sin(x))\cdot \cos(x)$

$=\cos( \sin( \sin ( \sin( \sin(x)))))\cdot \cos( \sin( \sin( \sin(x))))\cdot \cos( \sin( \sin(x)))\cdot \cos( \sin(x))\cdot \cos(x)$
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March 1st, 2017, 04:37 AM   #4
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Quote:
Originally Posted by nbg273 View Post
I'm trying to learn this stuff by myself. Can someone explain to me how to find the derivative of "sin(sin(sin(sin(sin(x)))))" using the Chain Rule? I'm confused by the whole concept.

Thank you.
Let u= sin(x). Then sin(sin(x))= sin(u). Also $\displaystyle \frac{du}{dx}= \cos(x)$.
Let v= sin(u). Then sin(sin(sin(x))= sin(v). Also $\displaystyle \frac{dv}{du}= \cos(u)= \cos(\sin(x))$.
Let w= sin(v). Then sin(sin(sin(sin(x)))= sin(w). Also $\displaystyle \frac{dw}{dv}= \cos(v)= \cos(\sin(u))= \cos(\sin(\sin(x))$.
Let y= sin(w). Then sin(sin(sin(sin(sin(x))))= sin(y). Also $\displaystyle \frac{dy}{dw}= \cos(y)= \cos(\sin(w))= \cos(\sin(\sin(v)))= \cos(\sin(\sin(u)))= \cos(\sin(\sin(\sin(x)))$.

So the derivative of sin(sin(sin(sin(sin(x)))) is the derivative of y with respect to x: $\displaystyle \frac{dy}{dx}= \frac{dy}{dw}\frac{dw}{dx}$

But $\displaystyle \frac{dw}{dx}=\frac{dw}{dv}\frac{dv}{dx}$ so $\displaystyle \frac{dy}{dx}= \frac{dy}{dw}\frac{dw}{dv}\frac{dv}{dx}$.

etc.
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Last edited by skipjack; March 1st, 2017 at 05:26 AM.
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March 2nd, 2017, 10:03 PM   #5
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Thanks, everyone!
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