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

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

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(\partial)/(\partial x)(1/2 ln(x^2+y^2))

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

\frac{x}{x ^{2} + y ^{2}}

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the chain rule:

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

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

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

= \frac{1}{2} \cdot \frac{1}{x ^{2} + y ^{2}} \cdot 2 x

Simplify

\frac{1}{2} \cdot \frac{1}{x ^{2} + y ^{2}} \cdot 2 x : \frac{x}{x ^{2} + y ^{2}}

= \frac{x}{x ^{2} + y ^{2}}

y = x^{4} + 4 e^{x}

\int (2 x sin (y) - y sin (x)) d x

\frac{d}{d x} ((8 x^{2} - 14 x) e^{x})

\int_{1}^{25} e^{x} d x

\frac{d}{d x} (ln ((x + 3)^{4}))

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