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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

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

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

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

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

Simplify

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

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

x \to - 4 lim (25)

2 y^{^{''}} + 7 y^{^{'}} - 4 y = 0

6 y^{^{'}} - 2 y = x (y^{4}), y (0) = - 2

y^{^{'}} - y^{(\frac{5}{3})} = 0

\frac{x ^{2}}{x ^{4} + 16}

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