What is (\partial)/(\partial x)((9xy^2)/(x^2+y^4)) ?

The answer to (\partial)/(\partial x)((9xy^2)/(x^2+y^4)) is (9y^2(-x^2+y^4))/((x^2+y^4)^2)

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(\partial)/(\partial x)((9xy^2)/(x^2+y^4))

\frac{\partial}{\partial x} (\frac{9 x y ^{2}}{x ^{2} + y ^{4}})

\frac{9 y ^{2} ( - x ^{2} + y ^{4} )}{( x ^{2} + y ^{4} ) ^{2}}

Solution steps

\frac{\partial}{\partial x} (\frac{9 x y ^{2}}{x ^{2} + y ^{4}})

Treat

y

as a constant

Take the constant out:

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

= 9 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}}

= 9 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

= 9 y^{2} \frac{1 \cdot ( x ^{2} + y ^{4} ) - 2 xx}{( x ^{2} + y ^{4} ) ^{2}}

Simplify

9 y^{2} \frac{1 \cdot ( x ^{2} + y ^{4} ) - 2 xx}{( x ^{2} + y ^{4} ) ^{2}} : \frac{9 y ^{2} ( - x ^{2} + y ^{4} )}{( x ^{2} + y ^{4} ) ^{2}}

= \frac{9 y ^{2} ( - x ^{2} + y ^{4} )}{( x ^{2} + y ^{4} ) ^{2}}

\frac{d}{d x} ((4 x^{2} + 3 x + 2)^{5})

\int_{0}^{2} \frac{6 x}{( 1 + x ^{2} ) ^{2}} d x

\frac{\partial}{\partial y} (y - x^{3})

\frac{d}{d x} (\frac{( 1 - x )}{x})

y^{^{''}} + 6 y^{^{'}} + 9 y = t^{- 2} e^{- 3 t}

What is (\partial)/(\partial x)((9xy^2)/(x^2+y^4)) ?
The answer to (\partial)/(\partial x)((9xy^2)/(x^2+y^4)) is (9y^2(-x^2+y^4))/((x^2+y^4)^2)