# Calculus problem

If $\lambda=\lambda\left(r\right)$ , how do you get from $$\left(1-2r\lambda'\right)e^{-2\lambda}=1$$ (the textbook then says, “integrates to”)$$\frac{d\left(re^{-2\lambda}\right)}{dr}=1$$

$$re^{-2\lambda}=r+c$$

I can see how you get from the second to the third equation, but not how to get from the first to the second? The prime means differentiated wrt r.

Many thanks.

-
You get the first from the second by differentiating, using the product/chain rule. – user20266 Dec 6 '11 at 20:25

Note that $$\frac{d}{dr}(re^{-2\lambda}) = e^{-2\lambda} + r\frac{d}{dr}e^{-2\lambda} = e^{-2\lambda} + r\left(e^{-2\lambda}\left(-2\lambda\right)'\right)$$ by the Product Rule and the Chain Rule. This gives the left hand side of the first equation. So you can replace the left hand side with $$\frac{d}{dr}(re^{-2\lambda}),$$ and then integrate to get the third equation.
 thanks, but how do you get from $\r\frac{d}{dr}e^{-2\lambda}\$ to $\r\left(e^{-2\lambda}\left(\frac{d}{dr}-2\lambda\right)'\right)\$ I know you said use the chain rule, but what do I call u? – Peter4075 Dec 6 '11 at 20:54 That's the chain rule: $u=-2\lambda$. You are trying to take the derivative of $e^u$ with respect to $r$, so the answer is $(e^u)u'$. – Arturo Magidin Dec 6 '11 at 20:56 Whoops. Thanks, but how do you get from $r\frac{d\left(e^{-2\lambda}\right)}{dr}$ to the final term? I know you said use the chain rule, but how exactly? In the format $\frac{dy}{dx}=\frac{du}{dx}\frac{dy} – Peter4075 Dec 6 '11 at 21:02 I said "how exactly":$y = e^{-2\lambda}$,$u=-2\lambda$, and you have no$x$s, you are differentiating with respect to$r$. – Arturo Magidin Dec 6 '11 at 21:03 I made a mess of the code and posted my "whoops" comment just as you were posting your "that's the chain rule" comment. I see it now. I couldn't figure out what$u$was supposed to represent. I seemed to want to differentiate wrt lambda and not$r\$, which added to my confusion. Many thanks for persevering. – Peter4075 Dec 7 '11 at 7:47