Differentials of Composite Functions

[LJoseph1227]

I cannot figure out how to do this problem completely:

If U =x³y, find \frac{dU}{dt} if x⁵ + y = t and x² + y³ = t².

I know that I am using the chain rule here and I have the partial derivates of U:
\frac{∂U}{∂x} = 3x²y

\frac{∂U}{∂y} = x³

So far I have the equation given below.
\frac{dU}{dt} = 3x²y \frac{dx}{dt} + x³ \frac{dy}{dt}

However, I do not know how to calculate \frac{dx}{dt} and \frac{dy}{dt}. I tried to calculate them implicitly but I am still working with three variables x, y, and t. Could you please help me with this? Any insight would be greatly appreciated! Thank you!

 

[A. Bahat]

Well, you haven't used either of the other conditions: x^5 + y = t and x^2 + y^3 = t^2. Try differentiating with respect to t. Then you can at least get \frac{dx}{dt} in terms of \frac{dy}{dt} (or vice versa).

 

[LJoseph1227]

Thank you for your suggestion! I really appreciate it! :)

Okay so I calculated \frac{dx}{dt} and \frac{dy}{dt}.

\frac{dx}{dt} = -\frac{\frac{∂F}{∂t}}{\frac{∂F}{∂x}} and \frac{dy}{dt} = -\frac{\frac{∂F}{∂t}}{\frac{∂F}{∂y}}

I substituted x and y into the two different t equations getting x² + (t - x⁵)³ - t² = 0 and (t² - y³)^5/2 + y - t = 0. From there I was able to get \frac{dx}{dt} = - [-2t + 3(t - x⁵)² / 2x - 15x⁴(t - x⁵)²] and \frac{dy}{dt} = - [5t(t² - y³)^\frac{3}{2} - 1 / -\frac{15}{2}y²(t² - y³)^\frac{3}{2} + 1].

From there I plugged \frac{dx}{dt} and \frac{dy}{dt} into the original equation, \frac{du}{dt} = \frac{∂u}{∂x} (\frac{dx}{dt}) + \frac{∂u}{∂y} (\frac{dy}{dt}).

The final answer for \frac{du}{dt} looks pretty ugly and cannot be simplified much. Does this seem correct? Is this how I ultimately calculate \frac{du}{dt}?