What is (\partial)/(\partial x)((ln(x^2+y^2))^{1/2}) ?

The answer to (\partial)/(\partial x)((ln(x^2+y^2))^{1/2}) is x/(ln^{1/2)(x^2+y^2)(x^2+y^2)}

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(\partial)/(\partial x)((ln(x^2+y^2))^{1/2})

\frac{\partial}{\partial x} ((ln (x^{2} + y^{2}))^{\frac{1}{2}})

\frac{x}{ln ^{\frac{1}{2}} ( x ^{2} + y ^{2} ) ( x ^{2} + y ^{2} )}

Solution steps

\frac{\partial}{\partial x} ((ln (x^{2} + y^{2}))^{\frac{1}{2}})

Treat

y

as a constant

Apply the chain rule:

\frac{1}{2 ln ^{\frac{1}{2}} ( x ^{2} + y ^{2} )} \frac{\partial}{\partial x} (ln (x^{2} + y^{2}))

= \frac{1}{2 ln ^{\frac{1}{2}} ( x ^{2} + y ^{2} )} \frac{\partial}{\partial x} (ln (x^{2} + y^{2}))

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

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

Simplify

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

= \frac{x}{ln ^{\frac{1}{2}} ( x ^{2} + y ^{2} ) ( x ^{2} + y ^{2} )}

\int (x - 2 y) d y

\frac{18}{x}

\int t^{- 4} d t

\frac{d}{d x} (\frac{x ^{2} + x - 2}{x ^{3} + 6})

y, ln (x y 9) = x y

What is (\partial)/(\partial x)((ln(x^2+y^2))^{1/2}) ?
The answer to (\partial)/(\partial x)((ln(x^2+y^2))^{1/2}) is x/(ln^{1/2)(x^2+y^2)(x^2+y^2)}