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

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

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

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

- 2 y

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

= 0 - 2 y

Simplify

= - 2 y

t a y l or ln (6 x + 1)

x \to \infty lim (\frac{e ^{5 x - 3}}{ln ( x - 2 )})

d er i v a t i v e f (x) = cos (a^{8} + x^{8})

\int_{0}^{\infty} x^{1} e^{- x} d x

\frac{d}{d x} (ln ((5 x^{2} + 9)^{3}))

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