Question

Let f(x,y)=xe^y -ye^x The gradient of f at (0,0) is ∇f(0,0)=?

Answer 1

`bbnabla f = (e^y - ye^x, xe^y - e^x)`

`bbnabla f(0,0) = (1,-1)`

Answer 2

`bb grad f(0,0) = bb hat(i) - bb hat(j)`

Explanation:

We have:

`f(x,y) = xe^y - ye^x`

We can compute the first order partial derivatives. Remember when partially differentiating that we differentiate wrt the variable in question whilst treating the other variables as constant. And so:

`f_x = (partial f)/(partial x)`
`\ \ \ = e^y - ye^x`

`f_y = (partial f)/(partial y)`
`\ \ \ = xe^y - e^x`

Then the Del, or gradient operator, using a standard basis, is given by:

`bb grad f(x,y) = (partial f)/(partial x) bb hat(i) + (partial f)/(partial y) bb hat(j)`

`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = (e^y - ye^x) bb hat(i) + (xe^y - e^x) bb hat(j)`

And so at the origin, we have:

`bb grad f(0,0) = (e^0 - 0) bb hat(i) + (0 - e^0) bb hat(j)`

`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = bb hat(i) - bb hat(j)`