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How do you find the derivative of #f(x)=1/(x-1)#?

Answer 1

Ezra Adams

#f'(x)=-(x-1)^-2#

#f(x)=(x-1)^-1#

#f'(x)=-1*(x-1)^(-1-1)*d/dx[x-1]# #color(white)(f'(x))=-(x-1)^-2#

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Answer 2

Emilia Aguilar

To find the derivative of ( f(x) = \frac{1}{x - 1} ), you can use the quotient rule. The quotient rule states that if you have a function in the form ( \frac{u(x)}{v(x)} ), then its derivative is given by:

[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} ]

Applying this rule to ( f(x) = \frac{1}{x - 1} ), where ( u(x) = 1 ) and ( v(x) = x - 1 ), we have:

[ f'(x) = \frac{(1)(x - 1)' - (1)(x - 1)'}{(x - 1)^2} ]

[ f'(x) = \frac{0 - 1}{(x - 1)^2} ]

[ f'(x) = -\frac{1}{(x - 1)^2} ]

So, the derivative of ( f(x) = \frac{1}{x - 1} ) is ( f'(x) = -\frac{1}{(x - 1)^2} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.