What is (\partial)/(\partial x)(1/(x^2+2y)) ?

The answer to (\partial)/(\partial x)(1/(x^2+2y)) is -(2x)/((x^2+2y)^2)

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(\partial)/(\partial x)(1/(x^2+2y))

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

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

Solution steps

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

Treat

y

as a constant

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

y = \frac{1}{2 x - 9}

\frac{d}{d x} (e^{- i x})

x \to 0 lim ((1 + x) \frac{1}{x})

\int \frac{15}{1 + 9 x ^{2}} d x

- 4 x + x^{2} + 10, 9 - 2 x, 2 x + 1

What is (\partial)/(\partial x)(1/(x^2+2y)) ?
The answer to (\partial)/(\partial x)(1/(x^2+2y)) is -(2x)/((x^2+2y)^2)