What is (\partial)/(\partial x)(3x^3y^3-2x^2+3y+40) ?

The answer to (\partial)/(\partial x)(3x^3y^3-2x^2+3y+40) is 9x^2y^3-4x

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(\partial)/(\partial x)(3x^3y^3-2x^2+3y+40)

\frac{\partial}{\partial x} (3 x^{3} y^{3} - 2 x^{2} + 3 y + 40)

9 x^{2} y^{3} - 4 x

Solution steps

\frac{\partial}{\partial x} (3 x^{3} y^{3} - 2 x^{2} + 3 y + 40)

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (3 x^{3} y^{3}) - \frac{\partial}{\partial x} (2 x^{2}) + \frac{\partial}{\partial x} (3 y) + \frac{\partial}{\partial x} (40)

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

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

\frac{\partial}{\partial x} (3 y) = 0

\frac{\partial}{\partial x} (40) = 0

= 9 y^{3} x^{2} - 4 x + 0 + 0

Simplify

= 9 x^{2} y^{3} - 4 x

2 - \frac{4}{t}

\int \frac{ln ( 13 x )}{x} d x

e^{3 x}

x \to \infty lim (\frac{e ^{2 x}}{5 ^{2 x}})

\frac{d}{d x} ((x^{2} + 1)^{4})

What is (\partial)/(\partial x)(3x^3y^3-2x^2+3y+40) ?
The answer to (\partial)/(\partial x)(3x^3y^3-2x^2+3y+40) is 9x^2y^3-4x