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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

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

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

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

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

Simplify

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

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

\frac{\partial}{\partial u} (2 cos (u))

y = - 2 x^{2} - 4 x + 1, (0, 1)

\frac{x}{2 x - 1}

\frac{d y}{d x} = x y + 3 x + 4 y + 12

\int ln (3 x - 2) d x

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