What is (\partial)/(\partial x)(-x^3-y^2+3x^2-10y-6) ?

The answer to (\partial)/(\partial x)(-x^3-y^2+3x^2-10y-6) is -3x^2+6x

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(\partial)/(\partial x)(-x^3-y^2+3x^2-10y-6)

\frac{\partial}{\partial x} (- x^{3} - y^{2} + 3 x^{2} - 10 y - 6)

- 3 x^{2} + 6 x

Solution steps

\frac{\partial}{\partial x} (- x^{3} - y^{2} + 3 x^{2} - 10 y - 6)

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} (y^{2}) + \frac{\partial}{\partial x} (3 x^{2}) - \frac{\partial}{\partial x} (10 y) - \frac{\partial}{\partial x} (6)

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

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

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

\frac{\partial}{\partial x} (10 y) = 0

\frac{\partial}{\partial x} (6) = 0

= - 3 x^{2} - 0 + 6 x - 0 - 0

Simplify

= - 3 x^{2} + 6 x

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

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

(x^{3} - 2 x + 1) (3 x^{3} + 2 x^{2} - 5 x)

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

\frac{( 3 y ^{2} - t ^{2} )}{y ^{5}} d y + \frac{t}{2 y ^{4}} d t = 0

What is (\partial)/(\partial x)(-x^3-y^2+3x^2-10y-6) ?
The answer to (\partial)/(\partial x)(-x^3-y^2+3x^2-10y-6) is -3x^2+6x