differentiation

You can use the product rule:

\(\displaystyle D_x[\frac{1}{x-6} \cdot \frac{1}{x-2}] = D_x[\frac{1}{x-6}](\frac{1}{x-2}) + (\frac{1}{x-6})D_x[\frac{1}{x-2}]\)

\(\displaystyle = \frac{-1}{(x-6)^2(x-2)} + \frac{-1}{(x-6)(x-2)^2} = \frac{-2x+8}{(x-6)^2(x-2)^2}\)


Or, you may multiply it out and use the chain rule:

\(\displaystyle D_x[\frac{1}{x-6} \cdot \frac{1}{x-2}] = D_x[(x^2-8x+12)^{-1}]\)

\(\displaystyle = -(x^2-8x+12)^{-2} \cdot (2x-8) = -\frac{2x-8}{(x^2-8x+12)^2}\)
 
Another similar but different representation of the same way:

\(\displaystyle D_x\left[\frac{1}{(x-6)(x-2)}\right ]\)

\(\displaystyle = \, \frac{1}{4} \cdot D_x\left[\frac{1}{(x-6)} - \frac{1}{(x-2)}\right ]\)

\(\displaystyle = \, \frac{1}{4} \cdot\left[\frac{1}{(x-2)^2} \, - \, \frac{1}{(x-6)^2}\right ]\)

Simplify further if needed.....
 
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