What is (\partial)/(\partial x)(-x^3+6x^2y^3+5y^2) ?

The answer to (\partial)/(\partial x)(-x^3+6x^2y^3+5y^2) is -3x^2+12xy^3

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(\partial)/(\partial x)(-x^3+6x^2y^3+5y^2)

\frac{\partial}{\partial x} (- x^{3} + 6 x^{2} y^{3} + 5 y^{2})

- 3 x^{2} + 12 x y^{3}

Solution steps

\frac{\partial}{\partial x} (- x^{3} + 6 x^{2} y^{3} + 5 y^{2})

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= - \frac{\partial}{\partial x} (x^{3}) + \frac{\partial}{\partial x} (6 x^{2} y^{3}) + \frac{\partial}{\partial x} (5 y^{2})

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

\frac{\partial}{\partial x} (6 x^{2} y^{3}) = 12 y^{3} x

\frac{\partial}{\partial x} (5 y^{2}) = 0

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

Simplify

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

\int \frac{2 x + 3}{( 2 x ^{2} + 6 x ) ^{4}} d x

\frac{d x}{d t} = \frac{1}{t ( 1 - t )}

\frac{d}{d t} (x y)

7 x sin (π x)

\int (x - \frac{1}{x ^{2}}) d x

What is (\partial)/(\partial x)(-x^3+6x^2y^3+5y^2) ?
The answer to (\partial)/(\partial x)(-x^3+6x^2y^3+5y^2) is -3x^2+12xy^3