Let y = (sin x)^3/4 (1 − cos x)^1/4 / x^3(√3x + 5). Use logarithmic differentiation to find y′.

Answer 1

In the statement of the problem, I am interpreting √3x as √(3x).

The rule of logarithmic differentiation is

y' = y (ln(y))' =

((sin x)^3/4 (1 − cos x)^1/4 / x³(√3x + 5)) (3/4 ln(sin(x)) + 1/4 ln(1-cosx) - 3ln(x) - ln(√3x + 5))' =

((sin x)^3/4 (1 − cos x)^1/4 / x³(√3x + 5)) (3/4 cos(x)/sin(x) + 1/4 sin(x)/(1-cosx) - 3/x - (√3/√x(√3x + 5))