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

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

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

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

2 y

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

= 2 y - 0

Simplify

= 2 y

\frac{d y}{d t} = y^{\frac{1}{3}} t^{\frac{2}{5}}

y^{^{'}} = 4 + e^{y - 4 x + 1}, y (0) = - 1

38 (e^{x^{2}} + 2 e^{x^{2}} x^{2})

\frac{0.5}{( s ^{2} + s + 1 ) \cdot s}

y = \frac{1}{1 - x}, x = 3

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