How do you differentiate #f(x)=(x-sinx)/(x-1)# using the quotient rule?

Question

Luke Alvarez · Accepted Answer

#f'(x)=(cosx(1-x)-1+sinx)/(x-1)^2# According to quotient rule for #f(x)=(p(x))/(q(x))# #f'(x)=(q(x)p'(x)-p(x)q'(x))/(q(x))^2# Hence, for #f(x)=(x-sinx)/(x-1)# #f'(x)=((x-1)(1-cosx)-(x-sinx)*1)/(x-1)^2# = #(x-xcosx-1+cosx-x+sinx)/(x-1)^2# = #(cosx(1-x)-1+sinx)/(x-1)^2#

Charles Arocha · Answer

undefined To differentiate $ f(x) = \frac{x - \sin(x)}{x - 1} $ using the quotient rule, follow these steps:

1. Identify the numerator $ u(x) = x - \sin(x) $ and the denominator $ v(x) = x - 1 $.
2. Apply the quotient rule formula: $ \frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{u'(x)v(x) - v'(x)u(x)}{(v(x))^2} $.
3. Find the derivatives of the numerator and denominator: $ u'(x) $ and $ v'(x) $.
4. Substitute the derivatives and original functions into the quotient rule formula.
5. Simplify the expression if necessary.

Here are the steps:

1. $ u(x) = x - \sin(x) $ and $ v(x) = x - 1 $.
2. $ u'(x) = 1 - \cos(x) $ and $ v'(x) = 1 $.
3. Apply the quotient rule formula: 
   $ f'(x) = \frac{(1 - \cos(x))(x - 1) - 1(x - \sin(x))}{(x - 1)^2} $.
4. Simplify the expression:
   $ f'(x) = \frac{x - \cos(x) - x + \sin(x) - (x - \sin(x))}{(x - 1)^2} $.
5. $ f'(x) = \frac{\sin(x) - \cos(x)}{(x - 1)^2} $.