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

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

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

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

- 6 y^{2}

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial y} (x^{4}) = 0

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

= 0 - 6 y^{2}

Simplify

= - 6 y^{2}

\frac{d}{d x} (x - 2 arctan (x))

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

\frac{d}{d x} (sin (4 x + 1))

cos (t) + t sin (t), at t = 0

\frac{d y}{d x} = \frac{x ^{2} + y ^{2} - x y}{x ^{2}}

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