Question #c5935

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Answer 1 · 2017-03-29T17:03:41

$d/dx (1/(x-1)) = -1/(x-1)^2$

Write the function as:

$1/(x-1) = (x-1)^-1$

Pose $y=(x-1)$ and differentiate using the chain rule:

$d/dx ( (x-1)^-1) = d/dy y^-1* dy/dx$

As:

$dy/dx = d/dx (x-1) = 1$

we get:

$d/dx ( (x-1)^-1) = d/dy y^-1$

Based on the power rule:

$d/dy y^n = ny^(n-1)$

we have:

$d/dx ( (x-1)^-1) = d/dy y^-1* = -1y^-2 = -1/y^2$

And substituting $y = x-1$ we can conclude:

$d/dx 1/(x-1)= -1/(x-1)^2$