# The chain rule with square root and $e^{3x}$

Alright so I am a little confused on how to proceed with this problem, I am supposed to find the derivative of $y= \sqrt {1+2e^{3x}}$ so I set the problem up like this $y=u^{1/2}$ $y = 1+2e^{3x}$ but then with that I have to use the chain rule again I believe on the term $2e^{3x}$ so I am not sure how to proceed from here, so I need to use the product rule on 2 and $e^{3x}$ or do I just consider 2e as one term since it is so simple? I can't remember having to use the chain rule on just a single integer like 2 before.

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If you have something like $2f(x)$, try applying the product rule to it, and remember the result so that you don't have to do it again. – J. M. Sep 24 '11 at 19:22

Think of applying the Chain Rule as an analogous process to opening a Matryoshka Doll. You cannot open all the dolls at the same time; you open the outermost doll, take out the next one. Then you open the second doll, and take out the third. Then you open the third, and take out the fourth, etc.

With $$y = \sqrt{1+2e^{3x}} = (1+2e^{3x})^{1/2}$$ the "outermost doll" is the square root. To "open it", we compute the derivative. Since $\frac{d}{du}(u^{1/2}) = \frac{1}{2}u^{-1/2}$, we take the derivative and then multiply by "the next doll": $$y' = \frac{1}{2}(1+2e^{3x})^{-1/2}(1+2e^{3x})'.$$ Then we need to take the derivative of $1+2e^{3x}$. By the sum rule, this is the derivative of $1$ plus the derivative of $2+e^{3x}$: $$y' = \frac{1}{2}(1+2e^{3x})^{-/12}\left( (1)' + (2e^{3x})'\right).$$ The derivative of the constant $1$ is $0$. The derivative of $2e^{3x}$ is $(2e^{3x})' = 2(e^{3x})'$: $$y' = \frac{1}{2}(1+2e^{3x})^{-1/2}\left( 0 + 2(e^{3x})'\right).$$ Finally, the derivative of $e^{3x}$. This is itself an application of the Chain Rule. We have $$\frac{d}{du}e^{u} = e^u,$$ so $$\frac{d}{dx}e^u = e^u\frac{du}{dx}.$$ That is: $$y' = \frac{1}{2}(1+2e^{3x})^{-1/2}\left( 0 + 2\left( e^{3x}(3x)'\right)\right).$$ And $(3x)' = 3$, so in the end we have: \begin{align*} y'&= \frac{1}{2}(1+2e^{3x})^{-1/2}\left( 0 + 2\left(e^{3x}(3)\right)\right) \\ &= \frac{1}{2}(1+2e^{3x})^{-1/2}\left(6e^{3x}\right) \\ &= 3e^{3x}(1+2e^{3x})^{-1/2}\\ &= \frac{3e^{3x}}{\sqrt{1+2e^{3x}}}. \end{align*}

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Thanks, I think I follow everything you did. I was able to get the proper answer. – user138246 Sep 24 '11 at 19:56

First tip is to separate out the functions. Like this: set $v = 3x$ and $u = e^v$, so then $y = (1 + u)^{1/2}$.

Now you use the chain rule:

$\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dv} \frac{dv}{dx}$

You should be able work out the three derivates on the right-hand side :)

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I did do that but just a little differently and in a way that made sense to me. Was there something wrong with how I split it up? – user138246 Sep 24 '11 at 19:22
You're using the chain rule on multiplication by constants, which isn't really that useful, since you should know how to differentiate that. You should the chain rule on more complicated functions; square-roots, exponents, elementary functions. – dmcdougall Sep 24 '11 at 19:36
So the derivative of 2e^3x is 2 (e^3x lne)? – user138246 Sep 24 '11 at 19:38
@Jordan: The derivative of $2e^{3x}$ is twice the derivative of $e^{3x}$. The derivative of $e^{3x}$ is $e^{3x}(3x)' = 3e^{3x}$, by the Chain Rule, using the fact that the derivative of $e^v$ (with respect to $v$) is $e^v$. – Arturo Magidin Sep 24 '11 at 19:40
So I can only do the ln thing when it is raised by a single digit? – user138246 Sep 24 '11 at 19:44

Work from the outside in. The outermost operation is the square root, so your first step is fine: you can let $u = 1 + 2e^{3x}$ and think of $y$ as $u^{1/2}$. Then from the power and chain rules you know that $y' = \frac12 y^{-1/2}u'$, so you need to calculate $u'$. You can do that term by term: you’ll need the derivative of $1$ and the derivative of $2e^{3x}$. It’s true that $2e^{3x}$ is a product, and you can use the product rule, but it’s silly to do so, because you know an easier way: if $v$ is any function of $x$, and $a$ is any constant, $(av)' = av'$. You wouldn’t use the product rule to differentiate $4x^3$, right? You’d just write down $4(3x^2)$ or $12x^2$. (You could: you’d get $4(x^3)' + (4)'x^3 = 4(3x^2) + 0\cdot x^3 = 12x^2$, which is right, but it’s a waste of energy and time.)

Note that you don’t have $2e$ anywhere in the expression: $2e^{3x}$ is $(2)(e^{3x})$, not $(2e)^{3x}$.

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So the derivative of $2(e)^{3x}$ should be $2(e^{3x} lne$)? – user138246 Sep 24 '11 at 19:28
@Jordan: The derivative of $a^{f(x)}$ is $a^{f(x)} f'(x) \ln(a)$, so the derivative of $e^{3x} = e^{3x} \cdot (3x)' \cdot \ln e$. But what is $\ln e$ anyway? – TMM Sep 24 '11 at 19:35
@Jordan: Not quite. It should be $2[e^{3x}]'$, but $[e^{3x}]'$ isn’t $e^{3x}\ln e$. In general $[a^u]'=e^u u'\ln a$, and $\ln e=1$, so $[e^u]'=e^u u'$. You forgot about the $u'$. – Brian M. Scott Sep 24 '11 at 19:36
I thought the forumla was $\prime{(a^x)} = a^x lna$ – user138246 Sep 24 '11 at 19:39
@Jordan: It is, when the exponent is the variable with respect to which you’re differentiating. When the exponent is a more complicated function of that variable, you must use the chain rule. Any time you differentiate a function that is a composition $-$ one function applied to the output of another $-$ you need the chain rule. $e^{3x}$ is such a function: first you apply the ‘multiply by $3$’ function, and then you exponentiate. Thus, $[e^{3x}]'=e^{3x}[3x]'$, not just $e^{3x}$. – Brian M. Scott Sep 24 '11 at 19:58

Let's define $u=1+2e^{3x}$ than we can write $y'=(\sqrt{u})'u'=\frac{u'}{2\sqrt{u}}=\frac{(1+2e^{3x})'}{2\sqrt{1+2e^{3x}}}=\frac{2(e^{3x})'}{2\sqrt{1+2e^{3x}}}= \frac{2e^{3x}(3x)'}{2\sqrt{1+2e^{3x}}}=\frac{6e^{3x}}{2\sqrt{1+2e^{3x}}}=\frac{3e^{3x}}{\sqrt{1+2e^{3x}}}$

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I am not used to this way of going through the problem, I will have to go over this for a long time to see what happened. I don't understand that notation at all or why you are using the derivative of u or 2 times the square root of u. – user138246 Sep 24 '11 at 19:27
@JordanCarlyon,en.wikipedia.org/wiki/… – pedja Sep 24 '11 at 19:43
Yeah I am not reading that, it is more confusing than my book and I have to study for this test I am going to fail on monday. – user138246 Sep 24 '11 at 19:44
$(u^{\frac{1}{2}})'=\frac{1}{2}u^{\frac{-1}{2}}$ – pedja Sep 24 '11 at 19:50
I know how to get a derivative. – user138246 Sep 24 '11 at 19:53