What is (\partial)/(\partial y)((2x)/(x^2+y^4)) ?

The answer to (\partial)/(\partial y)((2x)/(x^2+y^4)) is -(8xy^3)/((x^2+y^4)^2)

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(\partial)/(\partial y)((2x)/(x^2+y^4))

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

- \frac{8 x y ^{3}}{( x ^{2} + y ^{4} ) ^{2}}

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

\frac{\partial}{\partial y} (x^{2} + y^{4}) = 4 y^{3}

= 2 x (- \frac{1}{( x ^{2} + y ^{4} ) ^{2}} \cdot 4 y^{3})

Simplify

2 x (- \frac{1}{( x ^{2} + y ^{4} ) ^{2}} \cdot 4 y^{3}) : - \frac{8 x y ^{3}}{( x ^{2} + y ^{4} ) ^{2}}

= - \frac{8 x y ^{3}}{( x ^{2} + y ^{4} ) ^{2}}

f (x - π)

f (x) = tan (x) - 1

\frac{\partial}{\partial x} (5 e^{x} cos (yz))

\int_{0}^{1} (e^{- 2 x}) d x

\frac{d}{d x} (\frac{5}{8 x ^{8}})

What is (\partial)/(\partial y)((2x)/(x^2+y^4)) ?
The answer to (\partial)/(\partial y)((2x)/(x^2+y^4)) is -(8xy^3)/((x^2+y^4)^2)