What is (\partial)/(\partial y)(3x^2-2y^3+6xy^2-8) ?

The answer to (\partial)/(\partial y)(3x^2-2y^3+6xy^2-8) is -6y^2+12xy

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(\partial)/(\partial y)(3x^2-2y^3+6xy^2-8)

\frac{\partial}{\partial y} (3 x^{2} - 2 y^{3} + 6 x y^{2} - 8)

- 6 y^{2} + 12 x y

Solution steps

\frac{\partial}{\partial y} (3 x^{2} - 2 y^{3} + 6 x y^{2} - 8)

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial y} (3 x^{2}) - \frac{\partial}{\partial y} (2 y^{3}) + \frac{\partial}{\partial y} (6 x y^{2}) - \frac{\partial}{\partial y} (8)

\frac{\partial}{\partial y} (3 x^{2}) = 0

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

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

\frac{\partial}{\partial y} (8) = 0

= 0 - 6 y^{2} + 12 x y - 0

Simplify

= - 6 y^{2} + 12 x y

x^{2} 1, 8

f (x) = \frac{3 x}{x - 2}, at a = 4

x \to 4 lim (x - 2)

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

\int \frac{sin ^{2} ( x )}{1 + cos ( x )} d x

What is (\partial)/(\partial y)(3x^2-2y^3+6xy^2-8) ?
The answer to (\partial)/(\partial y)(3x^2-2y^3+6xy^2-8) is -6y^2+12xy