User Name Remember Me? Password

 Calculus Calculus Math Forum

 July 1st, 2017, 05:44 AM #1 Senior Member   Joined: Nov 2015 From: United States of America Posts: 198 Thanks: 25 Math Focus: Calculus and Physics Chain rule concept question Hello all, I have a quick question on the chain rule for derivatives. I have used it hundreds of times through out my study of calculus, but embarrassingly I get a little confused when, for example, you're doing a related rates problem and you take the derivative of a function you developed and then you multiply by a derivative after everything is said and done. Like: V = (1/3) pi r^3 dV/dt = (1/3)pi 3r^2 (dr/dt) I 100% understand that the when using the chain rule you take the derivative of the "outside" function, and then multiply by the derivative of the "inside" function. But when I am doing problems like this, I fail to see a composition of functions. And then I do not see the reason to utilize the chain rule. I would be forever thankful if someone could explain why this is the case. Thanks Jacob July 1st, 2017, 07:23 AM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 3,002 Thanks: 1587 Both $V$ and $r$ are implicit functions of time. Thanks from greg1313 July 1st, 2017, 02:06 PM #3 Member   Joined: Dec 2016 From: - Posts: 62 Thanks: 10 The easiest explanation is as follows: since $r$ depends on $t$, we have $r(t)$. The function $V(r)$ is then differentiated respect to time as: \begin{eqnarray} V(r(t))\to \frac{dV}{dt}=\frac{dV}{dr}\frac{dr}{dt} \end{eqnarray} and that is it. Just remember, for example, how to differentiate \begin{eqnarray} \sin(\log(x)) \end{eqnarray} we argue that $z=\log(x)$, so that the function $\sin(z)$ depends implicitly on the variable $x$ (see comment above). Then we all need to know is the formula I wrote above to differentiate, hope it helps now! Thanks from greg1313 Last edited by skipjack; July 2nd, 2017 at 02:28 AM. Tags chain, concept, question, rule Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post tweakedsam Calculus 1 February 23rd, 2013 06:41 AM unwisetome3 Calculus 4 October 19th, 2012 01:21 PM jasonperson Calculus 2 October 25th, 2011 04:09 AM Peter1107 Calculus 1 September 8th, 2011 10:25 AM David_Lete Calculus 4 October 21st, 2009 10:33 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top

Copyright © 2019 My Math Forum. All rights reserved.      