What is (\partial)/(\partial y)(4xe^{x^2+y^2}) ?

The answer to (\partial)/(\partial y)(4xe^{x^2+y^2}) is 8e^{x^2+y^2}xy

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(\partial)/(\partial y)(4xe^{x^2+y^2})

\frac{\partial}{\partial y} (4 x e^{x^{2} + y^{2}})

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

Solution steps

\frac{\partial}{\partial y} (4 x e^{x^{2} + y^{2}})

Treat

x

as a constant

Take the constant out:

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

= 4 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

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

Simplify

= 8 e^{x^{2} + y^{2}} x y

y^{^{'}} = (3 x - 2)^{2} (30 x + 29)

t y^{^{'}} + (t + 1) y = 2 t e^{- t}

\frac{2 s + 12}{s ^{2} + 6 s + 18}

\int \frac{5 x ^{3} - 4 x}{1 - x ^{4}} d x

\int_{0}^{1} \frac{x - 7}{x ^{2} - 5 x + 6} d x

What is (\partial)/(\partial y)(4xe^{x^2+y^2}) ?
The answer to (\partial)/(\partial y)(4xe^{x^2+y^2}) is 8e^{x^2+y^2}xy