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

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

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

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

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

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

Solution steps

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

Take the constant out:

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

= 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}

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

Simplify

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

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

t y^{^{'}} = (1 - y^{2})^{\frac{1}{2}}

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

\frac{1}{x - 1}, 18

sin (x), cos (x)

x \to 5 - lim (25 - x^{2})

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