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

The answer to (\partial)/(\partial x)(y/(1+x^2y^2)) is -(2y^3x)/((1+y^2x^2)^2)

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

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

- \frac{2 y ^{3} x}{( 1 + y ^{2} x ^{2} ) ^{2}}

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the chain rule:

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

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

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

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

Simplify

y (- \frac{1}{( 1 + x ^{2} y ^{2} ) ^{2}} \cdot 2 y^{2} x) : - \frac{2 y ^{3} x}{( 1 + y ^{2} x ^{2} ) ^{2}}

= - \frac{2 y ^{3} x}{( 1 + y ^{2} x ^{2} ) ^{2}}

y^{^{'}} - a \cdot y + b = 0

y^{^{'}} = \frac{( e ^{(\frac{x}{3})} + y )}{2}

\frac{d}{d x} (- 4 sin (8 x))

x \to 0 lim (\frac{x}{1 + x})

x^{2} - 6, 3

What is (\partial)/(\partial x)(y/(1+x^2y^2)) ?
The answer to (\partial)/(\partial x)(y/(1+x^2y^2)) is -(2y^3x)/((1+y^2x^2)^2)