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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Apply the Quotient Rule:

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

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

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

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

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

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

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

\int_{π}^{2 π} x d x

\int \frac{- 1}{1 - ( 4 t + 1 ) ^{2}} d t

e^{x^{2} - 1}

x \to 0 lim (ln (20))

\frac{d}{d x} (\frac{6 e ^{x}}{( 6 + e ^{x} ) ^{2}})

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