What is (\partial)/(\partial x)(ln(x+y/(2x))) ?

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

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(\partial)/(\partial x)(ln(x+y/(2x)))

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

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

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

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