What is (\partial)/(\partial x)((e^{xy})/(x+1)) ?

The answer to (\partial)/(\partial x)((e^{xy})/(x+1)) is (e^{xy}y(x+1)-e^{xy})/((x+1)^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\partial)/(\partial x)((e^{xy})/(x+1))

\frac{\partial}{\partial x} (\frac{e ^{x y}}{x + 1})

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

Solution steps

\frac{\partial}{\partial x} (\frac{e ^{x y}}{x + 1})

Treat

y

as a constant

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

\int u e^{- u} d u

y = 4 ln (x), at x = 4

\int_{0}^{1} (x^{\frac{4}{5}}) d x

\int \frac{( 1 + 6 ln ( x ) ) ^{11}}{x} d x

\int x^{2} + 2 x y + y^{2} d x

What is (\partial)/(\partial x)((e^{xy})/(x+1)) ?
The answer to (\partial)/(\partial x)((e^{xy})/(x+1)) is (e^{xy}y(x+1)-e^{xy})/((x+1)^2)