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

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

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

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

\int x^{2} e^{20 x} d x

n = 0 \sum \infty (x - 5)^{n}

f (x) = 2 π x

f (x) = x + 7, at x = 2

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