What is (\partial)/(\partial y)(3/2 (x+1)^2(y^2-9)) ?

The answer to (\partial)/(\partial y)(3/2 (x+1)^2(y^2-9)) is 3y(x+1)^2

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(\partial)/(\partial y)(3/2 (x+1)^2(y^2-9))

\frac{\partial}{\partial y} (\frac{3}{2} (x + 1)^{2} (y^{2} - 9))

3 y (x + 1)^{2}

Solution steps

\frac{\partial}{\partial y} (\frac{3}{2} (x + 1)^{2} (y^{2} - 9))

Treat

x

as a constant

Take the constant out:

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

= \frac{3}{2} (x + 1)^{2} \frac{\partial}{\partial y} (y^{2} - 9)

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{3}{2} (x + 1)^{2} (\frac{\partial}{\partial y} (y^{2}) - \frac{\partial}{\partial y} (9))

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

\frac{\partial}{\partial y} (9) = 0

= \frac{3}{2} (x + 1)^{2} (2 y - 0)

Simplify

\frac{3}{2} (x + 1)^{2} (2 y - 0) : 3 y (x + 1)^{2}

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

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What is (\partial)/(\partial y)(3/2 (x+1)^2(y^2-9)) ?
The answer to (\partial)/(\partial y)(3/2 (x+1)^2(y^2-9)) is 3y(x+1)^2