What is (\partial)/(\partial x)(3ln(x^2+y^2)) ?

The answer to (\partial)/(\partial x)(3ln(x^2+y^2)) is (6x)/(x^2+y^2)

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(\partial)/(\partial x)(3ln(x^2+y^2))

\frac{\partial}{\partial x} (3 ln (x^{2} + y^{2}))

\frac{6 x}{x ^{2} + y ^{2}}

Solution steps

\frac{\partial}{\partial x} (3 ln (x^{2} + y^{2}))

Treat

y

as a constant

Take the constant out:

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

= 3 \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

= 3 \cdot \frac{1}{x ^{2} + y ^{2}} \cdot 2 x

Simplify

3 \cdot \frac{1}{x ^{2} + y ^{2}} \cdot 2 x : \frac{6 x}{x ^{2} + y ^{2}}

= \frac{6 x}{x ^{2} + y ^{2}}

\frac{d}{d x} (sin (- 1 (2 x + 1)))

\int - sec^{2} (\frac{3 x}{2}) d x

\frac{d}{d x} (\frac{ln ( 2 x )}{x})

\int_{0}^{\frac{π}{2}} x \cdot sin (x) d x

\frac{d}{d x} (cos (2 ln (x)))

What is (\partial)/(\partial x)(3ln(x^2+y^2)) ?
The answer to (\partial)/(\partial x)(3ln(x^2+y^2)) is (6x)/(x^2+y^2)