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

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

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(\partial)/(\partial x)(-x/(1-x^2y^2))

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

= - \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}}

= - \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

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

Simplify

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

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

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

x \to \infty lim (\frac{1}{x} - 4)

f (x) = \frac{1 + x}{1 + e ^{x}}, at x = 0

\frac{10}{s ( s + 3 )}

2 x^{2} - 6 x

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