• عربي

Need Help?

Subscribe to Calculus C

###### \${selected_topic_name}
• Notes

Use the Chain Rule to find $d z / d t$
$z=x y^{3}-x^{2} y, \quad x=t^{2}+1, \quad y=t^{2}-1$

$z=x y^{3}-x^{2} y, \quad x=t^{2}+1, \quad y=t^{2}-1$

$\frac{d z}{d t}=\frac{\partial z}{\partial y} \frac{d y}{d t}+\frac{\partial z}{\partial x} \frac{d x}{d t}$

$\frac{d z}{d t}=\left[3 y^{2} x-x^{2}\right] (2 t)+\left[y^{3}-2 yx\right] (2 t)$

$\frac{d z}{d t}=2 t\left[3 y^{2} x-x^{2}+y^{3}-2 x y\right]$

Use the Chain Rule to find $\partial z / \partial s$ and $\partial z / \partial t$
$\quad z=(x-y)^{5}, \quad x=s^{2} t, \quad y=s t^{2}$

(1) $\frac{d z}{d t}=\frac{\partial z}{\partial y} \frac{\partial y}{\partial t}+\frac{\partial z}{\partial x} \frac{\partial x}{\partial t}$

$\frac{\partial z}{\partial t}=5(x-y)^{4}[-1](2 t s)+5(x-y)^{4}\left(s^{2}\right)=5(x-y)^{4} s^{2}-2 s t$

(2) $\frac{\partial z}{\partial s}=\frac{\partial z}{\partial y} \frac{\partial y}{\partial s}+\frac{\partial z}{\partial x} \frac{\partial x}{\partial s}$

$\frac{\partial z}{\partial s}=-5(x-y)^{4} t^{2}+5(x-y)^{4}(2 s t)$

$\frac{\partial z}{\partial s}=5(x-y)^{4}\left[2 s t-t^{2}\right]$