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

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

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

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

\frac{3 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}) = \frac{x}{x ^{2} + y ^{2}}

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

Simplify

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

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

\int \frac{x ^{3}}{x - 1} d x

(8 - 6 y + e^{- 3 x}) d x - 2 d y = 0

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

f (x) = e^{x} x^{- n}

\frac{d}{d x} ((3 x - x^{3} + 1)^{4})

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