What is (\partial)/(\partial x)((2xy^2)/(1+x^2y^2)) ?

The answer to (\partial)/(\partial x)((2xy^2)/(1+x^2y^2)) is (2y^2(1-x^2y^2))/((1+x^2y^2)^2)

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(\partial)/(\partial x)((2xy^2)/(1+x^2y^2))

\frac{\partial}{\partial x} (\frac{2 x y ^{2}}{1 + x ^{2} y ^{2}})

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

Solution steps

\frac{\partial}{\partial x} (\frac{2 x y ^{2}}{1 + 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}{1 + 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 ) ( 1 + x ^{2} y ^{2} ) - \frac{\partial}{\partial x} ( 1 + x ^{2} y ^{2} ) x}{( 1 + x ^{2} y ^{2} ) ^{2}}

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

\frac{\partial}{\partial x} (1 + x^{2} y^{2}) = 2 y^{2} x

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

Simplify

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

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

y = (x^{2} - 3) (x^{3} - 2 x), (- 2, - 4)

\frac{d}{d x} (\frac{lo g _{7} ( x - x ^{2} )}{x ^{3}})

[x^{2} (x^{2} + 5 x)]^{4}

e^{x^{4}}

x \to 3 lim ((x - 1)^{2})

What is (\partial)/(\partial x)((2xy^2)/(1+x^2y^2)) ?
The answer to (\partial)/(\partial x)((2xy^2)/(1+x^2y^2)) is (2y^2(1-x^2y^2))/((1+x^2y^2)^2)