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}}

\frac{d y}{d x} + \frac{1}{x} y = 0

\int 2 cos (\frac{x}{3}) d x

f (x) = x^{2} - 6 x - 7

x y^{2} y^{^{'}} = y^{3} - 5 x^{3}

\frac{d}{d θ} (tan^{2} (θ))

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}