Question

Sendo z=h(x,y)=x^2y-2y^2, x=2t e y=t^2-5, dz/dt é:

Answer 1

$\dfrac{dz}{dt}=\dfrac{\partial h}{\partial x}\cdot \dfrac{dx}{dt}+\dfrac{\partial h}{\partial y}\cdot \dfrac{dy}{dt}\\ \\ \\ \dfrac{dz}{dt}=(2xy)\cdot (2)+(x^2-4y)\cdot (2t)\\ \\ \\ \boxed{\dfrac{dz}{dt}=4xy+(x^2-4y)\cdot (2t)}$

                                 $\boxed{\dfrac{dz}{dt}=x^3}$