What is the d/(dy)(y^2e^{xy}) ?

The d/(dy)(y^2e^{xy}) is 2ye^{xy}+e^{xy}xy^2

What is the first d/(dy)(y^2e^{xy}) ?

The first d/(dy)(y^2e^{xy}) is 2ye^{xy}+e^{xy}xy^2

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

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

Solution steps

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

Treat

x

as a constant

Apply the Product Rule:

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

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

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

\frac{d}{d y} (y^{2}) = 2 y

\frac{d}{d y} (e^{x y}) = e^{x y} x

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

\frac{d}{d x} (x csc (x) - cot (x))

n = 0 \sum \infty \frac{6}{2 ^{n}}

f (x) = \frac{- 6}{x}

\frac{\partial}{\partial x} (- 8 x y - 2 x^{4} - 2 y^{4})

x \to 0 lim (\frac{6 ^{x} - 5 ^{x}}{x})

What is the d/(dy)(y^2e^{xy}) ?
The d/(dy)(y^2e^{xy}) is 2ye^{xy}+e^{xy}xy^2
What is the first d/(dy)(y^2e^{xy}) ?
The first d/(dy)(y^2e^{xy}) is 2ye^{xy}+e^{xy}xy^2