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

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

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

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

12 x^{3} y^{2}

Solution steps

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

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

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

= 12 x^{3} y^{2} - 0

Simplify

= 12 x^{3} y^{2}

\int \frac{4 + x + x ^{2}}{x} d x

y^{^{'}} = - \frac{y}{100}, y (0) = 80

\frac{d}{d x} (\frac{2}{cos ( x )})

\frac{d}{d x} (1 0^{x} + 10 x)

\int_{\frac{π}{2}}^{π} sin (2 x) d x

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