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find the derivative of loge(sinx) w.r.t loga(cosx)

let u = loge (sin x)
& v = loge (cos x)
we have to find du/ dv
du/dx = cot x
dv/dx = - tan x

du/dv = - cot2x
which is the required solution.
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let u = loge (sin x) and v = loga (cos x)
So we need to find du/dv
so, du/dx = d(loge (sin x))/dx = (1/sin x).(cos x) = cot x
and dv/dx = d(loge (cos x)/loge a)/dx = (1/loge a).d(loge (cos x))/dx = (1/loge a).(1/cos x).(-sin x) = -(tan x)/(loge a)
Therefore, du/dv=(du/dx)/(dv/dx) = (cot x)/(-(tan x)/(loge a)) = -(cot2 x).(loge a)
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let u = loge (sin x)
and v = loga (cos x)
So, we need to find du/dv
du/dx = d(loge (sin x))/dx = (1/sin x).(cos x) = cot x
dv/dx = d(loge (cos x)/loge (a))/dx = (1/loge a).(1/cos x).(-sin x) = -(tan x)/(loge a)

Therefore, du/dv = (du/dx)/(dv/dx) = (cot x​)/(-(tan x)/(loge a)) = -cot2 x.loge (a)

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