Question

How do you differentiate `xe^(x^2+y^2)`?

Answer 1

To differentiate a two-variable function, you need to build the gradient. The gradient is a vector with as many coordinates as the variables the function depends on.

Each coordinate of the vector is a derivative with respect to one of the variables. So, in the two-variables case, you need to calculate the derivatives with respect to `x` and `y`, and then put them together in a vector.

Since deriving with respect to a variable means to consider the other as a constant, it's easier to derivate your function if it's expressed in the form
`xe^{x^2+y^2}=x *e^{x^2} *e^{y^2}`

So, deriving with respect to `x`, and using the product rule `(fg)'=f'g+fg'` where `f(x)=x` and `g(x)=e^{x^2}`, we get
`d/dx x * e^{x^2} * e^{y^2} = e^{y^2} (e^{x^2} + x * e^{x^2} * 2x) =`
`e^{x^2} * e^{y^2} (1+2x^2)=e^{x^2+y^2} (1+2x^2)`
Where the derivative of `e^{x^2}` has been calculated using the chain rule, which states that `f(g(x))'=f'(g(x)) * g'(x)`, where `f(x)=e^x`, and `g(x)=x^2`.

The derivative with respect to `y` is easier, since the only factor to differentiate is `e^{y^2}`, while the others depend only on `x` and are thus to be considered as constant. So, we have
`d/dy x * e^{x^2} * e^{y^2} = x * e^{x^2} * e^{y^2} * 2y=`
`2xy e^{x^2+y^2}`

The gradient is thus the vector

`(e^{x^2+y^2} (1+2x^2), 2xy\ e^{x^2+y^2})`