Related Rates: Find $\frac{dr}{dt}$ given that $e^{xy}$...

Question

I have been trying to work this question out for a couple of days now and don't understand it. I know $r=e^{2×1/3}$ but can't work out how to do the derivatives with respect to $t$.

Suppose $\frac{dx}{dt}=4,\frac{dy}{dt}=8$. Find $\frac{dr}{dt}$ given that $r=e^{xy}$ when $x=6y=2$ exactly.

Answer 1

Hint: By the chain rule, $$ \frac{\mathrm dr}{\mathrm dt} = \frac{\mathrm d}{\mathrm dt} (e^{xy}) = e^{xy}\frac{\mathrm d}{\mathrm dt}(xy) = e^{xy}\left( \frac{\mathrm dx}{\mathrm dt} y + x\frac{\mathrm dy}{\mathrm dt} \right). $$

Answer 2

Notice that$${dr\over dt}={dr\over dx}{dx\over dt}+{dr\over dy}{dy\over dt}=4yr+8xr=e^{2\over 3}(16+{4\over 3})$$