What is (\partial)/(\partial x)((e^y)/(x+y^4)) ?

The answer to (\partial)/(\partial x)((e^y)/(x+y^4)) is -(e^y)/((x+y^4)^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^4))

\frac{\partial}{\partial x} (\frac{e ^{y}}{x + y ^{4}})

- \frac{e ^{y}}{( x + y ^{4} ) ^{2}}

Solution steps

\frac{\partial}{\partial x} (\frac{e ^{y}}{x + y ^{4}})

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 ^{4}})

Apply exponent rule:

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

= e^{y} \frac{\partial}{\partial x} ((x + y^{4})^{- 1})

Apply the chain rule:

- \frac{1}{( x + y ^{4} ) ^{2}} \frac{\partial}{\partial x} (x + y^{4})

= - \frac{1}{( x + y ^{4} ) ^{2}} \frac{\partial}{\partial x} (x + y^{4})

\frac{\partial}{\partial x} (x + y^{4}) = 1

= e^{y} (- \frac{1}{( x + y ^{4} ) ^{2}} \cdot 1)

Simplify

e^{y} (- \frac{1}{( x + y ^{4} ) ^{2}} \cdot 1) : - \frac{e ^{y}}{( x + y ^{4} ) ^{2}}

= - \frac{e ^{y}}{( x + y ^{4} ) ^{2}}

lo g_{b} (3)

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

y = - 6 x^{5} + 2

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

(9 x)^{c s c (6 x)}

What is (\partial)/(\partial x)((e^y)/(x+y^4)) ?
The answer to (\partial)/(\partial x)((e^y)/(x+y^4)) is -(e^y)/((x+y^4)^2)