What is (\partial)/(\partial x)(4xye^{xy^2}) ?

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

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(\partial)/(\partial x)(4xye^{xy^2})

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

= 4 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}}

= 4 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}

= 4 y (1 \cdot e^{x y^{2}} + e^{x y^{2}} y^{2} x)

Simplify

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

\int \frac{x ^{2}}{2} \cdot \frac{1}{x} d x

2 e^{- 5 t} sin (2 t)

y^{^{'}} = 3 3 y^{2}

f (x) = x ln (x^{2})

(- 2, - 4), (- 3, - 3)

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