What is (\partial)/(\partial y)(6xe^{x^2+y^2}) ?

The answer to (\partial)/(\partial y)(6xe^{x^2+y^2}) is 12e^{x^2+y^2}xy

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(\partial)/(\partial y)(6xe^{x^2+y^2})

\frac{\partial}{\partial y} (6 x e^{x^{2} + y^{2}})

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

Solution steps

\frac{\partial}{\partial y} (6 x e^{x^{2} + y^{2}})

Treat

x

as a constant

Take the constant out:

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

= 6 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

= 6 x e^{x^{2} + y^{2}} \cdot 2 y

Simplify

= 12 e^{x^{2} + y^{2}} x y

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

ln (x^{2} - 8 x + 1) (8)

\frac{d}{d x} (e^{x} - 2 sin (x))

\int (\frac{cot ^{2} ( x )}{cos ^{2} ( x )}) d x

\frac{( \frac{10}{6} )}{s ( s ^{2} + \frac{1}{2} )}

What is (\partial)/(\partial y)(6xe^{x^2+y^2}) ?
The answer to (\partial)/(\partial y)(6xe^{x^2+y^2}) is 12e^{x^2+y^2}xy