What is (\partial)/(\partial y)(x^{1/2}+y^{1/2}) ?

The answer to (\partial)/(\partial y)(x^{1/2}+y^{1/2}) is 1/(2y^{1/2)}

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(\partial)/(\partial y)(x^{1/2}+y^{1/2})

\frac{\partial}{\partial y} (x^{\frac{1}{2}} + y^{\frac{1}{2}})

\frac{1}{2 y ^{\frac{1}{2}}}

Solution steps

\frac{\partial}{\partial y} (x^{\frac{1}{2}} + y^{\frac{1}{2}})

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial y} (x^{\frac{1}{2}}) + \frac{\partial}{\partial y} (y^{\frac{1}{2}})

\frac{\partial}{\partial y} (x^{\frac{1}{2}}) = 0

\frac{\partial}{\partial y} (y^{\frac{1}{2}}) = \frac{1}{2 y ^{\frac{1}{2}}}

= 0 + \frac{1}{2 y ^{\frac{1}{2}}}

Simplify

= \frac{1}{2 y ^{\frac{1}{2}}}

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

\frac{u - 2}{u ^{2} - 2 u + 1} d u = - \frac{1}{x} d x

f (x) = (9 x - 1)^{- \frac{1}{3}}, at x = 1

t, \frac{d y}{d t} = y^{3} - y^{2} - 12 y

f (x) = \frac{4 e ^{x} + 5 e ^{- x}}{6}

What is (\partial)/(\partial y)(x^{1/2}+y^{1/2}) ?
The answer to (\partial)/(\partial y)(x^{1/2}+y^{1/2}) is 1/(2y^{1/2)}