Find $\frac{dz}{dt}$ if $z(x,y) = x^2y + xy^2$; $x(t) = 2 + t^4$; $y(t) = 1 - t^3$.

Question

Find $\frac{dz}{dt}$ if $z(x,y) = x^2y + xy^2$; $x(t) = 2 + t^4$; $y(t) = 1 - t^3$.

I used the Chain Rule: $\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt}$:

$$\frac{dz}{dt} = (2xy + y^2)(4t^3) + (x^2 + 2xy)(-3t^2)$$

I then substituted for the $x$'s and $y$'s to get the whole expression in terms of $t$'s and simplified to get

$$\frac{dz}{dt} = 8 -12t^2+ 6t^4 +12t^5- 14t^6 - 2t^7+t^8 + 10t^9 $$

Assuming this answer is correct, should my final answer be in terms of $t$ like this?

Answer 1

Yes, I would say a polynomial of $t$ is an acceptable answer. I think there is a small error in your calculations but other than that the method is correct. Real answer should be$$ \frac{dz}{dt}=t^2(-24+20t+12t^3-42t^4+8t^5+10t^7-11t^8). $$