What is (\partial)/(\partial y)(4x^2y^3-1) ?

The answer to (\partial)/(\partial y)(4x^2y^3-1) is 12x^2y^2

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(\partial)/(\partial y)(4x^2y^3-1)

\frac{\partial}{\partial y} (4 x^{2} y^{3} - 1)

12 x^{2} y^{2}

Solution steps

\frac{\partial}{\partial y} (4 x^{2} y^{3} - 1)

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial y} (4 x^{2} y^{3}) - \frac{\partial}{\partial y} (1)

\frac{\partial}{\partial y} (4 x^{2} y^{3}) = 12 x^{2} y^{2}

\frac{\partial}{\partial y} (1) = 0

= 12 x^{2} y^{2} - 0

Simplify

= 12 x^{2} y^{2}

\frac{d}{d x} (- 4 x^{2} - 2 x + 3)

7 (sin (x^{2}) + 2 x^{2} cos (x^{2}))

\frac{d}{d x} (5 x^{3} y^{2} + x^{2} sin (y^{2}))

\int \frac{11 x ^{4} + 11 x ^{3} + 2 x + 1}{x ^{2} + x} d x

\frac{d}{d x} (\frac{5 e ^{x} x ^{\frac{3}{2}}}{2} + e^{x} x^{\frac{5}{2}})

What is (\partial)/(\partial y)(4x^2y^3-1) ?
The answer to (\partial)/(\partial y)(4x^2y^3-1) is 12x^2y^2