Using the chain rule with a composite function I'm a little confused on this homework problem and I could use some explanation if anyone has seen something like it before.
The question is: Use the Chain Rule to find $\frac{dy}{dt}$ at $t = 9$ give then $y = \frac{u + 2}{u – 1}$, $u = (3s – 7)^2$, and $s = \sqrt{t}$.
I assume I'm not supposed to just plug in the values and take the derivative of the whole thing, as would be the intuitive solution. Instead I take the derivative of each part:
$$y' = -\frac{3}{(u-1)^2}$$
$$u' = 6(3s-7)$$
$$s' = \frac{1}{2\sqrt{t}}$$
However when I plug in 9 for t and follow the chain up I end up with $\frac{3}{774\sqrt{3} – 2092}$ which I highly doubt is the right answer. Could somebody help me out?
 A: 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:
$$\begin{align}t&=9\\s&=3\\u&=4\\y&=2\\&\dots\\\frac{dy}{dt}&=-\frac23\end{align}$$
A: Your differentiations are correct. If $t = 9$, then: $$\frac{\mathrm{d}s}{\mathrm{d}t}\Big|_{t = 9} = \frac{1}{6} \qquad \frac{\mathrm{d}u}{\mathrm{d}s}\Big|_{s = 3} = 12 \qquad \frac{\mathrm{d}y}{\mathrm{d}u}\Big|_{u = 4} = -\frac{1}{3}$$
Therefore: $$\frac{\mathrm{d}y}{\mathrm{d}t} = \frac{\mathrm{d}y}{\mathrm{d}u} \frac{\mathrm{d}u}{\mathrm{d}s} \frac{\mathrm{d}s}{\mathrm{d}t} = -\frac{2}{3}$$
I think you may have substituted the value os $s'$ and $u'$ instead of tha values of $s$ and $u$. When dealing with several variables, prime notation is not good, because it does not makes clear which variable are you differentiating with respect to.
