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#