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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x, g = e^{x^{2} + 2 y^{2}}

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

\frac{\partial}{\partial x} (x) = 1

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

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

Simplify

1 \cdot e^{x^{2} + 2 y^{2}} + e^{x^{2} + 2 y^{2}} \cdot 2 xx : e^{x^{2} + 2 y^{2}} + 2 e^{x^{2} + 2 y^{2}} x^{2}

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

f (x) = 3 x^{3} - x^{2} + 9, (2, 29)

\int \frac{x ln ( x )}{( 1 - x ^{2} ) ^{\frac{1}{2}}} d x

\frac{1}{( s ^{2} + 0.5 s + 1 )}

y^{^{''}} - 6 y^{^{'}} + 9 y = \frac{e ^{3 x}}{x ^{2}}

\frac{d y}{d x} = \frac{10 x ^{2}}{1 + x ^{3}}

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