Help with differential calculus?

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y = log(sinx)cosx = ln cos x / ln sin x
y ln sin x = ln cos x
y' ln sin x + y cos x / sin x = -sin x / cos x
y' ln sin x + (cos x / sin x)ln cos x / ln sin x = -sin x / cos x
y' ln sin x = -(cos x / sin x)ln cos x / ln sin x - sin x / cos x
y' ln sin x = -cot x log[sin x] cos x - tan x
y' = -(cot x log[sin x] cos x + tan x) / ln sin x...Show more

y = log(sinx) cos x
(sinx) ^y = cos x

y log sin x = log cos x
y = log cos x / log sin x
dy/dx = ((log sin x )(-sinx )/(cos x) - (log cos x) (cos x /sin x))/ (log sin x ) ^2

now you can simplify...Show more