partial derivatives: z = x/y, x = se^t, y = 1 + se^(-t)

[mathstresser]

Use the chain rule to find dz/ds and dz/dt.

z=x/y
x=se^t
y=1+se^(-t)

I know that

dz/ds=(dz/dx)(dx/ds)+(dz/dy)(dy/ds)
and
dz/dt=(dz/dx)(dx/dt)+(dz/dy)(dy/dt)

My problem is dz/dy.

The reason I have a problem is becuase I'm not exactly sure what to do with it becuase it's in the denominator. I would normally be able to do it, but I'm not sure what to do since it is a partial derivative.

Is it (yx-x)/(y^2) ? or (y-x)/(y^2)?

 

[skeeter]

\(\displaystyle \L z = \frac{x}{y}\)
\(\displaystyle \L x = se^t\)
\(\displaystyle \L y = 1 + se^{-t}\)

\(\displaystyle \L \frac{\partial z}{\partial x} = \frac{1}{y}\)

\(\displaystyle \L \frac{\partial z}{\partial y} = \frac{-x}{y^2}\)

\(\displaystyle \L \frac{\partial x}{\partial s} = e^t\)

\(\displaystyle \L \frac{\partial y}{\partial s} = e^{-t}\)

\(\displaystyle \L \frac{\partial x}{\partial t} = se^t\)

\(\displaystyle \L \frac{\partial y}{\partial t} = -se^{-t}\)

 

[mathstresser]

Thanks!