What is (\partial)/(\partial y)(2xe^{x^2+y^2}) ?

The answer to (\partial)/(\partial y)(2xe^{x^2+y^2}) is 4e^{x^2+y^2}xy

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(\partial)/(\partial y)(2xe^{x^2+y^2})

\frac{\partial}{\partial y} (2 x e^{x^{2} + y^{2}})

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

Solution steps

\frac{\partial}{\partial y} (2 x e^{x^{2} + y^{2}})

Treat

x

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the chain rule:

e^{x^{2} + y^{2}} \frac{\partial}{\partial y} (x^{2} + y^{2})

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

\frac{\partial}{\partial y} (x^{2} + y^{2}) = 2 y

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

Simplify

= 4 e^{x^{2} + y^{2}} x y

\frac{\partial}{\partial x} ((x y - 2) e^{- x y})

\frac{( ( 2 x + 5 ) ^{7} )}{14}

x^{4} - 8 x^{2} + 16

\frac{d}{d x} (\frac{- 3}{x - 6} - 2)

\frac{d}{d x} (0.8 \cdot x^{4})

What is (\partial)/(\partial y)(2xe^{x^2+y^2}) ?
The answer to (\partial)/(\partial y)(2xe^{x^2+y^2}) is 4e^{x^2+y^2}xy