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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

= 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}}

= 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

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

Simplify

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

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

\frac{\partial}{\partial v} (2 sin (u) sin (v))

\frac{d}{d x} (\frac{5 x}{x + 2})

\int 5 e^{x} - \frac{1}{x} d x

sin (2 arccos (x))

\int_{0}^{\frac{π}{18}} cos (6 x) d x

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