1. ## Partial derivative

Find $\frac{dw}{ds}(-2,2)$ and $\frac{dw}{dt}(-2,2)$ if $w= xy -yz -5xz,$ where $x = st, y = e^{st}, z= t^2$

I'm confused since only two coordinates are given but the problem has 3 variables $x,y, z$

2. Originally Posted by chaos787
Find $\frac{dw}{ds}(-2,2)$ and $\frac{dw}{dt}(-2,2)$ if $w= xy -yz -5xz,$ where $x = st, y = e^{st}, z= t^2$

I'm confused since only two coordinates are given but the problem has 3 variables $x,y, z$
You need to apply the chain rule: $\frac{\partial w}{\partial s}=\frac{\partial w}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial w}{\partial y}\frac{\partial y}{\partial s}+\frac{\partial w}{\partial z}\frac{\partial z}{\partial s}$

Similarly, $\frac{\partial w}{\partial t}=\frac{\partial w}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial w}{\partial y}\frac{\partial y}{\partial t}+\frac{\partial w}{\partial z}\frac{\partial z}{\partial t}$

Can you take it from here?

3. Originally Posted by chaos787
Find $\frac{dw}{ds}(-2,2)$ and $\frac{dw}{dt}(-2,2)$ if $w= xy -yz -5xz,$ where $x = st, y = e^{st}, z= t^2$

I'm confused since only two coordinates are given but the problem has 3 variables $x,y, z$
The "two coordinates" are values of s and t. When s= -2, t= 2, x= (-2)(2)= -4, $y= e^{(-2)(2)}= e^{-4}$, and z= 2^2= 4. Actually, it is a little ambiguous as to which of "(-2, 2)" is s and which t but it turns out that, in this problem, it doesn't matter!