partial derivative

Question

Find d(f(x,y))/ dt where: f(x,y)=e^ xy3 with x = t ^ 2 + s, y = t ^ 3 + s ^ 2 , and s, t > 0 .

Answer 1

If you want to find the partial derivative df/dt then your best bet is to apply the chain rule. With respect to t, your chain rule would be: df/dt = (df/dx)(dx/dt) + (df/dy)(dy/dt). I'm assuming your original function is f(x,y) = e^{xy^3}.

In which case df/dx = e^{xy^3}(y³) (normal chain rule with respect to x) and dx/dt = 2t

Also df/dy = e^{xy^3}(3xy²) (normal chain rule with respect to y) and dy/dt = 3t²

Putting all the pieces together you have df/dt = e^{xy^3}(y³)(2t) + e^{xy^3}(3xy²)(3t²).

If you'd like to have your final answer only in terms of s and t(which is probably what is requested), then you can substitute for x and y accordingly.