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.
[Math] The chain rule with square root and $e^{3x}$
calculus
Related Solutions
You may use any method that works. A convenient feature of mathematics is that it works no matter which (valid) method you use.
For this kind of problem, it is easiest to ask yourself, "how many 'places' is there an $x$?" By "places", I mean "numerator and denominator". If there are $x$s in both places, then you (probably) will prefer to use the quotient rule. If there are $x$s only in the numerator, then there is no quotient and the result is straightforward. If there are $x$s only in the denominator, then the quotient rule will be very easy; the lack of $x$s in the numerator makes the derivative of the numerator especially simple. Similarly, the chain rule makes it easy in the case that there are only $x$s in the denominator since this is "some function of $x$" raised to "some negative power".
Consider $g(x) = \frac{1}{f(x)}$. We may apply the quotient rule: $$g'(x) = \frac{f(x) \cdot 0 - 1 \cdot f'(x)}{f(x)^2} = \frac{-f'(x)}{f(x)^2}$$ or we may apply the chain rule: $$g'(x) = \left( f(x)^{-1} \right)' = (-1)f(x)^{-2}f'(x) = \frac{-f'(x)}{f(x)^2}$$. We get the same answer either way, so it doesn't matter which method we use.
Some people find it easier to go one way or the other. It doesn't matter which way is chosen as long as the method is applied correctly.
People don't normally write this as the chain rule, but in reality it is:
$$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$$
Now let's see what you have. You found the derivatives correctly: $$\begin{align}y' &= -\frac{3}{(u-1)^2}\\ u' &= 6(3s-7)\\ s' &= \frac{1}{2\sqrt{t}}\end{align}$$
Because the chain rule is algebraic, it's easier to see if I write the derivatives in the more technical notation to visualize what you took the derivative against (i.e. $y'=x\Rightarrow \frac{dy}{dx}=x$):
$$\begin{align}\frac{dy}{du} &= -\frac{3}{(u-1)^2}\\ \frac{du}{ds} &= 6(3s-7)\\ \frac{ds}{dt} &= \frac{1}{2\sqrt{t}}\end{align}$$
How would you get to $\frac{dy}{dt}$? Well, you'd use the above and do a little dimensional analysis. First start with the $\frac{dy}{du}$. Now you need to cancel the $du$, so multiply by $\frac{du}{ds}$. Need to cancel the $ds$ too, so multiply by $\frac{ds}{dt}$. Now you are left with $\frac{dy}{dt}$ and it was pretty easy. So again, we have:
$$\frac{dy}{dt}=\frac{dy}{du}\frac{du}{ds}\frac{ds}{dt}=\left(-\frac{3}{(u-1)^2}\right)\left(6(3s-7)\right)\left(\frac{1}{2\sqrt{t}}\right)$$
Now you need to evaluate it at $t=9$. Plug $9$ into the original $s$ function to find $s$. Using $s$, find $u$ from its function. Likewise, find $y$ using $u$ and you've got all the variables you need.
If my math is correct, you have:
$$\begin{align}t&=9\\s&=3\\u&=4\\y&=2\\&\dots\\\frac{dy}{dt}&=-\frac23\end{align}$$
Best Answer
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: $$\begin{align*} y'&= \frac{1}{2}(1+2e^{3x})^{-1/2}\left( 0 + 2\left(e^{3x}(3)\right)\right) \\ &= \frac{1}{2}(1+2e^{3x})^{-1/2}\left(6e^{3x}\right) \\ &= 3e^{3x}(1+2e^{3x})^{-1/2}\\ &= \frac{3e^{3x}}{\sqrt{1+2e^{3x}}}. \end{align*}$$