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

The answer to (\partial)/(\partial x)(2ln(x^2+y^2)) is (4x)/(x^2+y^2)

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

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

\frac{4 x}{x ^{2} + y ^{2}}

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

= 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

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

Simplify

2 \cdot \frac{1}{x ^{2} + y ^{2}} \cdot 2 x : \frac{4 x}{x ^{2} + y ^{2}}

= \frac{4 x}{x ^{2} + y ^{2}}

arccsc (- t^{2})

x \to 0 lim (sin (\frac{3 π}{x}))

\frac{\partial}{\partial x} (e^{x + t})

\frac{\partial}{\partial x} (6 e^{x y})

\frac{d}{d x} (e^{x} - x - 1)

What is (\partial)/(\partial x)(2ln(x^2+y^2)) ?
The answer to (\partial)/(\partial x)(2ln(x^2+y^2)) is (4x)/(x^2+y^2)