# Show the negative derivative of a function.

A type of interaction between atoms in a molecule is called a Van der Waals interaction. This can be described by the potential energy function; $$U= U_{0}\left(\frac{R_{0}}{r}\right)^{12}-2\left(\frac{R_{0}}{r}\right)^{6}$$

Also, the two atoms are distance $R_0$ from one another when in equilibrium. Let $x$ be the displacement from this equilibrium position so that, $x=r-{R_{0}} , r={R_{0}}+x$ Where the constant $U_{0}>0$.

So, we can find the $\text{force}(F)$ on the second atom by finding the negative derivative of $U$.

Show that the negative derivative of $U$ is

$$F= 12 \frac{U_{0}}{R_{0}}\left[\left(\frac{R_{0}}{r}\right)^{13}-\left(\frac{R_{0}}{r}\right)^{7}\right].$$

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So you have to take a derivative here. What do you know about derivatives? –  mixedmath Sep 30 '12 at 6:10
In particular, this would appear to be a (negative) derivative with respect to $r$. –  Gerry Myerson Sep 30 '12 at 7:22
@John: $F = - \dfrac{dU}{dr}$ is the relationship between potential energy and force here. The minus sign in front of the derivative is why your problem states "negative derivative." So the problem's just asking you to compute $- \dfrac{dU}{dr}$ and show you end up with the expression for $F$ that is given. Are there parentheses missing from the definition of $U$? –  Jonas Meyer Oct 5 '12 at 6:39
No, thats how it was given. –  John Oct 5 '12 at 12:47

U can be rewritten as $$U = U_0R_0^{12} \left( 1\over r\right)^{12} - 2R_0^{6} \left( 1\over r\right)^{6}$$ Now, $$F = - \frac{dU}{dr}$$ $$= -\frac{d}{dr}\left( U_0R_0^{12} \left( 1\over r\right)^{12} - 2R_0^{6} \left( 1\over r\right)^{6} \right)$$ $$= -\left( U_0R_0^{12} \frac{d}{dr}r^{-12} - 2R_0^{6} \frac{d}{dr}r^{-6} \right)$$ $$= -\left(U_0R_0^{12} (-12r^{-13}) - 2R_0^{6} (-6r^{-7}) \right)$$ Multiple and divide by $R_0$ $$= -\frac1{R_0}\left(U_0R_0^{13} (-12r^{-13}) - 2R_0^{7} (-6r^{-7}) \right)$$ $$= \frac{12}{R_0}\left(U_0R_0^{13} r^{-13} - R_0^{7} (r^{-7}) \right)$$ This equals to $$F = \frac{12}{R_0}\left[U_0\left(\frac{R_0}{r} \right)^{13} - \left(\frac{R_0}{r} \right)^{7}\right]$$ instead of $$F = 12\frac{U_0}{R_0}\left[\left(\frac{R_0}{r} \right)^{13} - \left(\frac{R_0}{r} \right)^{7}\right]$$ $U_0$ is not the coefficient of the second term.

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ok Thanks Shashwat, but i dont see how the second last equation equals to the last one. Could you clear that up? –  John Oct 5 '12 at 6:26
thanks mate i really appreciate it. –  John Oct 5 '12 at 6:46
But it doesn't equals to the one which you gave. –  Shashwat Oct 5 '12 at 7:49
But it must right? –  John Oct 5 '12 at 12:27
Then the answer can be wrong. You can see the $U_0$ is not the coefficient of the second term, so it can't be taken out as common as in the expression on $F$. –  Shashwat Oct 8 '12 at 4:40

Hint: $\left( \frac c x\right)^n=c^n\cdot x^{-n}$

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sorry its been so long since answering, I lost this site. embarrassing, I know. I still dont know what is meant by negative derative and this hint doesnt help me much. Sorry, but could someone help me with the steps? –  John Oct 5 '12 at 5:17

$$F={\bf \color{red}-}\frac{dU}{dr}$$${ }$ ${ }$${ }$

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The important thing to realize in solving this problem is that the potential energy function can be expressed in terms of either $x$ or $r$ depending on how you choose to express it from the relation $x = r - R_0$. In either case, note that $U$ is a composition of functions. $U$ is a function of $x$ which in turn is a function of $r$.