Question

differeniate Sin^cosx^tanx + logcotx.
Answer in detail​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

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Given, (cos x)y = (sin y)x

Take log on both side, we get

y * log cos x = x * log sin y

Now, differentiate both side w.r.t. x, we get

(dy/dx) * log cos x + y * (-sin x/cos x)= log sin y + (cos y/sin y) * (dy/dx)

=> (dy/dx) * [log cos x - cot y] = log sin y + y * tan x

=> dy/dx = (log sin y + y * tan x)/(log cos x -

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