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

The derivative of y=ln(x^2+y^2) is (2x)/(x^2+y^2)

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

The first derivative of y=ln(x^2+y^2) is (2x)/(x^2+y^2)

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derivative of

y = ln (x^{2} + y^{2})

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

Solution steps

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

Treat

y

as a constant

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

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

Simplify

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

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

\frac{\partial}{\partial y} (e^{x^{2} - y^{2}})

\frac{d}{d x} (1.5 (\frac{1}{3} + \frac{x}{2} - \frac{x ^{3}}{6}))

\int \frac{1}{- x ^{2} - 2 x 8} d x

\int \frac{( 1 )}{x ^{4}} d x

(x^{2})^{^{'}}

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