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

The answer to (\partial)/(\partial y)(x^2-y/4) is -1/4

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

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

- \frac{1}{4}

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

\frac{\partial}{\partial y} (\frac{y}{4}) = \frac{1}{4}

= 0 - \frac{1}{4}

Simplify

= - \frac{1}{4}

9

f (x) = \frac{8 x}{x ^{2} + 1}, at x = 0

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

\frac{25 x ^{3}}{3} + 10 x^{2} + 4 x - 7

f (x) = \frac{2}{x + 3}

What is (\partial)/(\partial y)(x^2-y/4) ?
The answer to (\partial)/(\partial y)(x^2-y/4) is -1/4