use the chain rule to find $\frac{dz}{dt}$, where $z = x^{2}y+xy^{2}$, $x = 4 - t^{3}$, $y = 3 - t^{5}$…

Explanation:

Step1: Find $\frac{\partial z}{\partial x}$

Differentiate $z = x^{2}y+xy^{2}$ with respect to $x$ treating $y$ as a constant. Using the power - rule for differentiation, we get $\frac{\partial z}{\partial x}=2xy + y^{2}$.

Step2: Find $\frac{\partial z}{\partial y}$

Differentiate $z = x^{2}y+xy^{2}$ with respect to $y$ treating $x$ as a constant. Using the power - rule for differentiation, we get $\frac{\partial z}{\partial y}=x^{2}+2xy$.

Step3: Find $\frac{dx}{dt}$

Differentiate $x = 4 - t^{3}$ with respect to $t$. Using the power - rule, $\frac{dx}{dt}=-3t^{2}$.

Step4: Find $\frac{dy}{dt}$

Differentiate $y = 3 - t^{5}$ with respect to $t$. Using the power - rule, $\frac{dy}{dt}=-5t^{4}$.

Step5: Apply the chain rule

The chain rule states that $\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}$. Substitute the values we found above: [ \begin{align*} \frac{dz}{dt}&=(2xy + y^{2})(-3t^{2})+(x^{2}+2xy)(-5t^{4})\ \end{align*} ] Since $x = 4 - t^{3}$ and $y = 3 - t^{5}$, we substitute $x$ and $y$ into the above expression: [ \begin{align*} \frac{dz}{dt}&=\left2(4 - t^{3})(3 - t^{5})+(3 - t^{5})^{2}\right+\left(4 - t^{3})^{2}+2(4 - t^{3})(3 - t^{5})\right \end{align*} ]

Answer:

[ \begin{align*} \frac{dz}{dt}&=\left2(4 - t^{3})(3 - t^{5})+(3 - t^{5})^{2}\right+\left(4 - t^{3})^{2}+2(4 - t^{3})(3 - t^{5})\right \end{align*} ]