What is (\partial)/(\partial x)(2xye^{xy^2}) ?

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

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(\partial)/(\partial x)(2xye^{xy^2})

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the Product Rule:

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

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

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

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

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

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

Simplify

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

f (x) = tan (x), (\frac{π}{4}, 1)

\int \frac{4}{x ^{2} + 2 x + 5} d x

y (1 + x^{2}) y^{^{'}} - x (8 + y^{2}) = 0

\frac{d}{d x} (\frac{6}{x} - \frac{8}{x})

a^{x}

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