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

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

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

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

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

Solution steps

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

Treat

y

as a constant

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

Simplify

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

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

y^{^{''}} = 4 y

\int tan (2 x) (sec (2 x))^{2} d x

\frac{s + 12}{( s ^{2} + 4 s )}

\frac{d}{d t} (\frac{5}{( 5 - t ) ^{2}})

\int \frac{x - 5}{x ^{3} + 2 x ^{2} + x} d x

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