What is the second derivative?

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

1 Answer

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