What is (\partial)/(\partial x)((e^y)/(x+y^8)) ?

The answer to (\partial)/(\partial x)((e^y)/(x+y^8)) is -(e^y)/((x+y^8)^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^y)/(x+y^8))

\frac{\partial}{\partial x} (\frac{e ^{y}}{x + y ^{8}})

- \frac{e ^{y}}{( x + y ^{8} ) ^{2}}

Solution steps

\frac{\partial}{\partial x} (\frac{e ^{y}}{x + y ^{8}})

Treat

y

as a constant

Take the constant out:

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

= e^{y} \frac{\partial}{\partial x} (\frac{1}{x + y ^{8}})

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= e^{y} \frac{\partial}{\partial x} ((x + y^{8})^{- 1})

Apply the chain rule:

- \frac{1}{( x + y ^{8} ) ^{2}} \frac{\partial}{\partial x} (x + y^{8})

= - \frac{1}{( x + y ^{8} ) ^{2}} \frac{\partial}{\partial x} (x + y^{8})

\frac{\partial}{\partial x} (x + y^{8}) = 1

= e^{y} (- \frac{1}{( x + y ^{8} ) ^{2}} \cdot 1)

Simplify

e^{y} (- \frac{1}{( x + y ^{8} ) ^{2}} \cdot 1) : - \frac{e ^{y}}{( x + y ^{8} ) ^{2}}

= - \frac{e ^{y}}{( x + y ^{8} ) ^{2}}

f (x) = x^{2} - 12, g (x) = 4 x

\frac{d}{d x} (\frac{cos ( x ) - 1}{x})

y^{^{''}} - 12 y + 36 y = 0

\frac{( e ^{- 1 - x} - e ^{- 1} )}{x}

\int_{1}^{3} \frac{5}{x ^{2} + 2 x + 1} d x

What is (\partial)/(\partial x)((e^y)/(x+y^8)) ?
The answer to (\partial)/(\partial x)((e^y)/(x+y^8)) is -(e^y)/((x+y^8)^2)