Question

derivative of (cosx-cos2x) /(1-cosx)​

Answer 1

$y = \frac{ \cos \: x - \cos \: 2x}{1 - \cos \: x } \\ \\ y = \frac{cosx - (2 {cos}^{2} x - 1)}{1 - cosx} \\ \\ y = \frac{2 {cos}^{2}x - cosx - 1 }{cosx - 1} \\ \\ y = \frac{2 {cos}^{2}x - 2cosx + cosx - 1 }{cosx - 1} \\ \\ y = \frac{2cosx(cosx - 1) + 1(cosx - 1)}{cosx - 1} \\ \\ y = \frac{(2cosx + 1)(cosx - 1)}{(cosx - 1)} \\ \\ y = 2cosx + 1 \\ \\ differentiate \: \: w.r.t .\: \: x \\ \\ \frac{dy}{dx} = - 2sinx$

hope it helps

Answer 2

Answer :

d/dx {(cosx - cos2x)/(1 - cosx)}

= - 2 sinx

Solution :

Now, cosx - cos2x

= cosx - (2 cos²x - 1)

= cosx - 2 cos²x + 1

= 1 + cosx - 2 cos²x

= 1 + 2 cosx - cosx - 2 cos²x

= 1 (1 + 2 cosx) - cosx (1 + 2 cosx)

= (1 + 2 cosx) (1 - cosx)

Then, d/dx {(cosx - cos2x)/(1 - cosx)}

= d/dx {(1 + 2 cosx) (1 - cosx)}/(1 - cosx)

= d/dx (1 + 2 cosx)

= d/dx (1) + 2 d/dx (cosx)

= - 2 sinx

Rules :

d/dx (cosx) = - sinx

d/dx (sinx) = cosx