Suppose v(s,t)=tsf(s+e^2t , s-e^(-3t)) where f(x,y) is some scalar valued function. Find dv/dt.

I think this is partial derivatives, but I'm not sure what it means when they ask tsf of the x,y value...

Please explain clearly.

Since v(s,t) = ts * f(x, y), where x = s+e^(2t) and y = s-e^(-3t),

∂v/∂t = s * f(x,y) + ts * (∂/∂t) f(x,y), by the product rule

........= s * f + ts * (∂f/∂x ∂x/∂t + ∂f/∂y ∂y/∂t), by the Chain Rule

........= s * f + ts * (∂f/∂x * (2e^(2t)) + ∂f/∂y * 3e^(-3t)).

----------

Could you explain the second line?

∂v/∂t = s * f(x,y) + ts * (∂/∂t) f(x,y), by the product rule

And

s * f + ts * (∂f/∂x * (2e^(2t)) + ∂f/∂y * 3e^(-3t)).

What do you do with the ∂f/∂x and ∂f/∂y after?