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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

n = 0 \sum \infty \frac{1}{ln ( n + 2 )}

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

x \to + 3 lim (2 x^{2} - 3 x + 4)

\int \frac{1}{x ^{\frac{5}{3}}} d x

\frac{\partial}{\partial x} (200 (x - y^{2}))

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