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

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

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

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

x

Solution steps

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

Treat

y

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Sum/Difference Rule:

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

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

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

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

= \frac{1}{2} (2 x + 0)

Simplify

\frac{1}{2} (2 x + 0) : x

= x

y^{^{''}} - \frac{2}{x} y^{^{'}} - \frac{10}{x ^{2}} y = 0

cos (4 x)

3 x + 4, at x = 4

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

f (x) = e^{x} x + 2 e^{x}

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