What is (\partial)/(\partial y)(x^5-6x+4y^3-y^2) ?

The answer to (\partial)/(\partial y)(x^5-6x+4y^3-y^2) is 12y^2-2y

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(\partial)/(\partial y)(x^5-6x+4y^3-y^2)

\frac{\partial}{\partial y} (x^{5} - 6 x + 4 y^{3} - y^{2})

12 y^{2} - 2 y

Solution steps

\frac{\partial}{\partial y} (x^{5} - 6 x + 4 y^{3} - y^{2})

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial y} (x^{5}) - \frac{\partial}{\partial y} (6 x) + \frac{\partial}{\partial y} (4 y^{3}) - \frac{\partial}{\partial y} (y^{2})

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

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

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

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

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

Simplify

= 12 y^{2} - 2 y

6 x^{4} - 3

\int x^{4} (x^{5} - 4)^{8} d x

f (x) = \frac{3}{3 x + 1}

\frac{d}{d x} (\frac{x ^{5} + 2 x + 6}{x ^{\frac{1}{3}}})

What is (\partial)/(\partial y)(x^5-6x+4y^3-y^2) ?
The answer to (\partial)/(\partial y)(x^5-6x+4y^3-y^2) is 12y^2-2y