What is the derivative of 1/(2ln(x^2+y^2)) ?

The derivative of 1/(2ln(x^2+y^2)) is -(x+y\frac{dy)/(dx)}{ln^2(x^2+y^2)(x^2+y^2)}

What is the first derivative of 1/(2ln(x^2+y^2)) ?

The first derivative of 1/(2ln(x^2+y^2)) is -(x+y\frac{dy)/(dx)}{ln^2(x^2+y^2)(x^2+y^2)}

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

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

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

What is the derivative of 1/(2ln(x^2+y^2)) ?
The derivative of 1/(2ln(x^2+y^2)) is -(x+y\frac{dy)/(dx)}{ln^2(x^2+y^2)(x^2+y^2)}
What is the first derivative of 1/(2ln(x^2+y^2)) ?
The first derivative of 1/(2ln(x^2+y^2)) is -(x+y\frac{dy)/(dx)}{ln^2(x^2+y^2)(x^2+y^2)}