What is the derivative of 1/(x+y) ?

The derivative of 1/(x+y) is -(1+\frac{dy)/(dx)}{(x+y)^2}

What is the first derivative of 1/(x+y) ?

The first derivative of 1/(x+y) is -(1+\frac{dy)/(dx)}{(x+y)^2}

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

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

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int 6 x + 12 x ln (8 x) d x

\int (x^{- 2}) d x

(10 y^{2} - x y) d x + x^{2} d y = 0

x \to \infty lim (6 x^{2})

n = 0 \sum \infty \frac{1}{1 + ln ( n )}

What is the derivative of 1/(x+y) ?
The derivative of 1/(x+y) is -(1+\frac{dy)/(dx)}{(x+y)^2}
What is the first derivative of 1/(x+y) ?
The first derivative of 1/(x+y) is -(1+\frac{dy)/(dx)}{(x+y)^2}