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: 2,379 Thanks: 1278 
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!


Tags 
chain, rule 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
chain rule constant rule  ungeheuer  Calculus  1  July 30th, 2013 05:10 PM 
Chain Rule?  FireExit  Calculus  4  February 8th, 2013 08:23 AM 
Product rule into chain rule  unwisetome3  Calculus  4  October 19th, 2012 01:21 PM 
how do yo solve this one using chain rule with quotient rule  Peter1107  Calculus  1  September 8th, 2011 10:25 AM 
chain rule  stainsoftime  Calculus  2  December 14th, 2008 11:00 PM 