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.
The chain rule with square root and $e^{3x}$
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0If 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. – 2011-09-24
4 Answers
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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0Thanks, I think I follow everything you did. I was able to get the proper answer. – 2011-09-24
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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0Ok thanks, I get that. – 2011-09-24
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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0So I can onl$y$ do the ln thing when it is raised by a single digit? – 2011-09-24
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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0I know how to get a derivative. – 2011-09-24