What is the d/(dy)(1/(x^2+y^2)) ?

The d/(dy)(1/(x^2+y^2)) is -(2y)/((x^2+y^2)^2)

What is the first d/(dy)(1/(x^2+y^2)) ?

The first d/(dy)(1/(x^2+y^2)) is -(2y)/((x^2+y^2)^2)

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

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

Solution steps

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

Treat

x

as a constant

Apply exponent rule:

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

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

Apply the chain rule:

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

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

Simplify

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

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

x^{2} - 4, x + 2

x \to 0 lim (- x ln (x))

\int sin (2 x) - cos (3 x) d x

y^{^{''}} + y = x^{2}

y^{2} = 4 x, y = 2 x - 4

What is the d/(dy)(1/(x^2+y^2)) ?
The d/(dy)(1/(x^2+y^2)) is -(2y)/((x^2+y^2)^2)
What is the first d/(dy)(1/(x^2+y^2)) ?
The first d/(dy)(1/(x^2+y^2)) is -(2y)/((x^2+y^2)^2)