Question

What is the second derivative?

Find the second derivative of `sinx/(1-cosx)`

Answer 1

Second derivative `f''(x)color(green)( = sin x / (cosx -1)^2)`

Explanation:

If the two functions u and v are differentiable (i.e. the derivative exist) then the quotient is differentiable and,

`(u/v)' = ((vu' - u v') / v^2)`

`f(x) = sin x / (1-cosx)`

`u = sinx, v = 1-cosx`

`u' = cos x, v' = sin x`

`f'(x) = (((1-cosx)cos x)- (sinx* sin x)) / (1-cosx)^2`

`f'(x) = (cos x - cos^2 x - sin^2 x) / (1-cos x)^2`

`f'(x) = (cos x - 1) / (1 - co^2 x) = (-1)/ (1-cos x)`

`f'(x) = -(1-cos x)^-1`

`f''(x) = cancel(-) (cancel(-1)(1-cos x)^-2) * sin x`

`f''(x)= sin x / (1-cos x)^2`