July 1st, 2017, 05:44 AM  #1 
Senior Member Joined: Nov 2015 From: United States of America Posts: 167 Thanks: 21 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: 2,755 Thanks: 1405 
Both $V$ and $r$ are implicit functions of time.

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! Last edited by skipjack; July 2nd, 2017 at 02:28 AM. 

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