# Differentials of Composite Functions

1. Jan 18, 2012

### 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!

2. Jan 18, 2012

### 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).

3. Jan 23, 2012

### 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 $\frac{du}{dt}$?

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook