What is (\partial)/(\partial x)(0.5ln(x^2+y^2)) ?

The answer to (\partial)/(\partial x)(0.5ln(x^2+y^2)) is x/(x^2+y^2)

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(\partial)/(\partial x)(0.5ln(x^2+y^2))

\frac{\partial}{\partial x} (0.5 ln (x^{2} + y^{2}))

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

Solution steps

\frac{\partial}{\partial x} (0.5 ln (x^{2} + y^{2}))

Treat

y

as a constant

Take the constant out:

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

= 0.5 \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

= 0.5 \cdot \frac{1}{x ^{2} + y ^{2}} \cdot 2 x

Simplify

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

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

y = 5 x^{6} ln (x)

x \to 0 lim (\frac{1}{cos ( x ) - 1})

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

\int \frac{8 t - 5}{t + 1} d t

What is (\partial)/(\partial x)(0.5ln(x^2+y^2)) ?
The answer to (\partial)/(\partial x)(0.5ln(x^2+y^2)) is x/(x^2+y^2)