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)

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(\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}}

x \to - 1 lim (x^{2} - 4)

\frac{d}{d x} (\frac{x + 6}{x ^{2} + 6})

\frac{d}{d x} (3 x^{2} - 5 x + 6)

\frac{d y}{d x} = - 8 y

\int 17 x^{16} d 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)