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October 1st, 2017, 04:24 PM  #1 
Newbie Joined: Oct 2017 From: ... Posts: 5 Thanks: 0  How do you find V prime of r for this function?
I'm really confused about how I am meant to solve this question. I'm assuming that I need to use the quotient rule but because everything involved is a variable I'm really confused.

October 1st, 2017, 05:15 PM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,940 Thanks: 794 
Every thing is not a variable! (or, more grammatically, "not everything is a variable") "r" is the only variable. The problem specifically says that $\displaystyle \epsilon$ and R are constants. The function is $\displaystyle V(r)= \epsilon\left(\left(\frac{R}{r}\right)^{12} 2\left(\frac{R}{r}\right)^6\right)$ If you let $\displaystyle x= \frac{R}{r}$ then $\displaystyle V= \epsilon\left(x^{12} x^6\right)$ so that $\displaystyle \frac{dV}{dx}= \epsilon\left(12x^{11} 12x^5\right)$. By the chain rule, $\displaystyle \frac{dV}{dr}= \frac{dV}{dx}\frac{dx}{dr}$. The derivative of $\displaystyle x= \frac{R}{r}$, with respect to r, using the quotient rule, is $\displaystyle \frac{dx}{dr}\frac{0(r) R(1)}{r^2}= \frac{R}{r^2}$. So $\displaystyle \frac{dV}{dr}= V'= \epsilon\left(12\left(\frac{R}{r}\right)^{11} 12\left(\frac{R}{r}\right)^5\right)\left(\frac{R}{r^2}\right)$. By the way you might find it easier to deal with fractions by writing the denominator in the numerator with a negative power. For example, here, $\displaystyle V(r)= \epsilon\left(R^{12}r^{12} 2R^6r^{6}\right)$ and then, just using the power formula, $\displaystyle V'= \epsilon\left(12R^{12}r^{13}+ 12R^6r^{7}\right)$. With a little algebra, you can show that this is the same answer as before. Last edited by Country Boy; October 1st, 2017 at 05:44 PM. 
October 1st, 2017, 05:36 PM  #3 
Newbie Joined: Oct 2017 From: ... Posts: 5 Thanks: 0 
Wow, thanks so much, it makes a lot more sense now. I should probably practice more though. Thanks again!


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