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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the Product Rule:

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

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

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

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

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

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

Simplify

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

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

\frac{d}{d x} (2 x^{4} + 4 x^{3} + x)

f (x) = \frac{x ^{2}}{1 + 2 x}

x \to 0 lim (1 - \frac{x ^{2}}{6})

\frac{t d u}{d t} = t^{2} + 3 u, t > 0, H (5) = 100

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