How do you differentiate x e x 2 + y 2

Question

How do you differentiate       x          e                        x          2                +                  y          2

tun1t2j · Accepted Answer

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

x e^{x^{2} + y^{2}} = x \cdot e^{x^{2}} \cdot e^{y^{2}}

So, deriving with respect to x, and using the product rule

(f g)' = f' g + f g'

where

f (x) = x

and

g (x) = e^{x^{2}}

, we get

\frac{d}{d x} x \cdot e^{x^{2}} \cdot e^{y^{2}} = e^{y^{2}} (e^{x^{2}} + x \cdot e^{x^{2}} \cdot 2 x) =

e^{x^{2}} \cdot e^{y^{2}} (1 + 2 x^{2}) = e^{x^{2} + y^{2}} (1 + 2 x^{2})

Where the derivative of

e^{x^{2}}

has been calculated using the chain rule, which states that

f (g (x))' = f' (g (x)) \cdot 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 consideblack as constant. So, we have

\frac{d}{d y} x \cdot e^{x^{2}} \cdot e^{y^{2}} = x \cdot e^{x^{2}} \cdot e^{y^{2}} \cdot 2 y =

2 x y e^{x^{2} + y^{2}}

The gradient is thus the vector

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

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

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