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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

\frac{\partial}{\partial x} (x) = 1

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

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

Simplify

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

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

\int \frac{7 x + 4}{x ^{3} + 4 x ^{2}} d x

y^{^{'}} + 3 (tan (3 x)) y = 7 cos (3 x)

x \to 1 lim ((1 - x)^{l n (x)})

\int_{1}^{5} 4 x^{2} ln (x) d x

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