What is the d/(dy)(ln(x^2+y^2)) ?

The d/(dy)(ln(x^2+y^2)) is (2y)/(x^2+y^2)

What is the first d/(dy)(ln(x^2+y^2)) ?

The first d/(dy)(ln(x^2+y^2)) is (2y)/(x^2+y^2)

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

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

Solution steps

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

Treat

x

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d}{d x} (\frac{6}{5} e^{- 20 x})

ln (\frac{x}{2}), x = 2

x \to - 1 lim (x^{5} - 1)

\frac{d}{d x} (\frac{24 x ( - x ^{2} + 1 )}{( x ^{2} + 1 ) ^{4}})

y^{^{'}} = 0.1 ln (\frac{4000}{y}) y, y (0) = 500

What is the d/(dy)(ln(x^2+y^2)) ?
The d/(dy)(ln(x^2+y^2)) is (2y)/(x^2+y^2)
What is the first d/(dy)(ln(x^2+y^2)) ?
The first d/(dy)(ln(x^2+y^2)) is (2y)/(x^2+y^2)