February 27th, 2017, 10:13 AM  #1 
Member Joined: Feb 2017 From: East U.S. Posts: 40 Thanks: 0  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. 
February 27th, 2017, 10:33 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,530 Thanks: 1390 
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? 
March 1st, 2017, 01:21 AM  #3 
Member Joined: Sep 2016 From: India Posts: 88 Thanks: 30 
According to @rosmek solution next steps will be $=\cos(\sin(\sin(\sin(x))))\cdot \cos(\sin(\sin(x)))\cdot\cos( \sin(x))\cdot \cos(x)$ $=\cos( \sin( \sin ( \sin( \sin(x)))))\cdot \cos( \sin( \sin( \sin(x))))\cdot \cos( \sin( \sin(x)))\cdot \cos( \sin(x))\cdot \cos(x)$ 
March 1st, 2017, 04:37 AM  #4  
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902  Quote:
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. Last edited by skipjack; March 1st, 2017 at 05:26 AM.  
March 2nd, 2017, 10:03 PM  #5 
Member Joined: Feb 2017 From: East U.S. Posts: 40 Thanks: 0 
Thanks, everyone!


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