chain rule for multi-variables The monthly demand for the Instant Pie Maker is given by
$$D(x,y)= \frac1{125}xe^{xy/1000} \text{ units}$$
where  $x$ dollars are spent on infomercials and $y$ dollars are spent on in-person demonstrations. If $t$ months from now $$x=20+t^{2/3}$$ dollars are spent on infomercials and $$y=t\ln(1+t)$$ dollars are spent on demonstrations, at approximately what rate will the demand be changing with respect to time $8$ months from now?
(Round your answer to $3$ decimal places).
I started this problem by finding $dz/dx=$ $$\frac{e^{xy/1000}(xy+1000)}{125000}$$
and multiplying it by $dx/dt$
$$\frac{2}{3t^{1/3}}$$
and repeated for $y$ getting $$\frac{x^2 e^{xy/1000}}{125000}\ln(t+1)+\frac{t}{t+1}$$
finally I plugged in $$t=8, x=24, y=8\ln 9$$
and I got $.526$ which is wrong. can anyone help me see where I went wrong or what i might be missing here?
 A: If I understood your problem correctly, then you have to first plug $x = 20 + t^{\frac{2}{3}}$ and $y = t\ln(1 + t)$ into $D(x,y)$ and then calculate it's derivative with respect to $t$ at the point $t = 8$. Here is a possible solution:
First we define
\begin{equation}
\begin{split}
\tilde D(t) := D(20 + t^{\frac{2}{3}}, t\ln(1+t)) &= \frac{1}{125}(20 + t^{\frac{2}{3}})\text{e}^\frac{(20t + t^{\frac{5}{3}})\ln(1 + t)}{1000}\\ &= \frac{1}{125}(20\text{e}^\frac{(20t + t^{\frac{5}{3}})\ln(1 + t)}{1000} + t^{\frac{2}{3}}\text{e}^\frac{(20t + t^{\frac{5}{3}})\ln(1 + t)}{1000})
\end{split}
\end{equation}
Next we want to calculate it's derivative. To keep things simple, we set
\begin{equation}
g(t) := \text{e}^\frac{(20t + t^{\frac{5}{3}})\ln(1 + t)}{1000} 
\end{equation}
Because
\begin{equation}
\frac{\text{d}}{\text{d}t} \frac{(20t + t^{\frac{5}{3}})\ln(1 + t)}{1000} = \frac{1}{1000}((20 + \frac{5}{3}t^{\frac{2}{3}})\ln(1 + t) + \frac{20t + t^{\frac{5}{3}}}{1+t}) 
\end{equation}
we have
\begin{equation}
g'(t) =\frac{1}{1000}((20 + \frac{5}{3}t^{\frac{2}{3}})\ln(1 + t) + \frac{20t + t^{\frac{5}{3}}}{1+t})g(t)
\end{equation}
Finally, we get 
\begin{equation}
\tilde D'(t) = \frac{1}{125}(20g'(t) + \frac{2}{3}t^{-\frac{1}{3}}g(t) + t^{\frac{2}{3}}g'(t))= \frac{1}{125}((20 + t^{\frac{2}{3}})g'(t) + \frac{2}{3}t^{-\frac{1}{3}}g(t))
\end{equation}
All we have to do now is to evaluate $\tilde D'(t)$ at $t = 8$, and according to Mathematica this is 
\begin{equation}
\tilde D'(8) = 0.0274655 \approx 0.027
\end{equation}
