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

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

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(\partial)/(\partial y)((1+x^2)/(3y-y^2))

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

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

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

\frac{\partial}{\partial y} (3 y - y^{2}) = 3 - 2 y

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

Simplify

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

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

\frac{d}{d j} (2 i + 3 j)

x^{2} - 4

(x^{2} + 1)^{^{'}}

\int \frac{e ^{x + 2}}{e ^{x + 1}} d x

18 cos (θ) + 9 sin^{2} (θ)

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