Answer to Question #136786 in Calculus for Promise Omiponle

Question

Use the chain rule to find dzdt, where
z=x^2y+xy^2,x=−4+t^2,y=5−t^7

First the pieces:

∂z/∂x=

∂z/∂y=

dx/dt=

dy/dt=


End result (in terms of just t):
dz/dt=

Answer 1

$z=x^2y+xy^2, x=−4+t^2, y=5−t^7\\ \displaystyle\frac{\partial z}{\partial x} = 2xy + y^2, \frac{\partial z}{\partial y} = x^2 + 2xy\\ \frac{\mathrm{d}x}{\mathrm{d}t} = 2t, \frac{\mathrm{d}y}{\mathrm{d}t} = -7t^6\\ \frac{\mathrm{d}z}{\mathrm{d}t} = \frac{\partial z}{\partial x} \cdot \frac{\mathrm{d}x}{\mathrm{d}t} + \frac{\partial z}{\partial y} \cdot\frac{\mathrm{d}y}{\mathrm{d}t}\\ \begin{aligned} \frac{\mathrm{d}z}{\mathrm{d}t} &= (2xy + y^2).2t + (2xy + x^2).(-7t^6) \\&= (2xy + y^2).2t + (2xy + x^2).(-7t^6) \\&= 4xyt + 2y^2t - 14t^6xy - 7x^2 t^6 \\&=4(5−t^7)(−4+t^2)t + 2(5−t^7)^2 t -\\& 14t^6(5−t^7)(−4+t^2) - 7(-4+t^2)^2 t^6 \\&= 2(5−t^7)^2 t + 2(5−t^7)(−4+t^2)(2t - 7t^6) - \\&7(-4+t^2)^2 t^6 \end{aligned}$