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October 1st, 2017, 03:24 PM   #1
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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.
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October 1st, 2017, 04:15 PM   #2
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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 04:44 PM.
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October 1st, 2017, 04:36 PM   #3
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Wow, thanks so much, it makes a lot more sense now. I should probably practice more though. Thanks again!
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