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

May 7, 2016

$f ' \left(x\right) = \frac{\cos x \left(1 - x\right) - 1 + \sin x}{x - 1} ^ 2$

Explanation:

According to quotient rule for $f \left(x\right) = \frac{p \left(x\right)}{q \left(x\right)}$

$f ' \left(x\right) = \frac{q \left(x\right) p ' \left(x\right) - p \left(x\right) q ' \left(x\right)}{q \left(x\right)} ^ 2$

Hence, for $f \left(x\right) = \frac{x - \sin x}{x - 1}$

$f ' \left(x\right) = \frac{\left(x - 1\right) \left(1 - \cos x\right) - \left(x - \sin x\right) \cdot 1}{x - 1} ^ 2$

= $\frac{x - x \cos x - 1 + \cos x - x + \sin x}{x - 1} ^ 2$

= $\frac{\cos x \left(1 - x\right) - 1 + \sin x}{x - 1} ^ 2$