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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

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

\frac{d}{d x} (- 1 - 4 sin (x))

k = 0 \sum \infty 5 x^{k}

\frac{s + 5}{s ^{2} - 4 s + 4}

\frac{d}{d x} (sin (2 arcsec (x)))

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