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

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

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(\partial)/(\partial y)((e^{xy})/(x+y))

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

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

Solution steps

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

Treat

x

as a constant

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

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

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

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

Simplify

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

\int \frac{1}{y ^{2} + y + 1} d y

\frac{d}{d x} (e^{l n (c o s (x^{2}))} 2 x + 1)

\frac{d y}{d x} + 2 x y = 5 x

(- 4, - 6), (- 2, 0)

n = 1 \sum \infty n^{5} (1 + n^{6})^{6}

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