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

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

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

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

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

Solution steps

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

Treat

t

as a constant

Take the constant out:

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

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

Apply exponent rule:

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

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

Simplify

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

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

\frac{4 x ^{2} + 6 x + 6}{x}

x \to \frac{π}{2} lim (\frac{1}{cos ( x )})

\frac{d}{d x} (\frac{cos ( 3 ) x ^{6}}{2 x})

x \to \infty + lim (2 + \frac{1}{x})

\int \frac{2 x ^{3} - x}{x ^{4} - x ^{2} - 25} d x

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