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Question
Find `bb(dy/dx)` for the given function:
(cos x)y = (cos y)x
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Solution
Given, (cos x)y = (cos y)x
Taking logarithm of both sides,
log (cos x)y = log (cos y)x
y log cos x = x log cos y
Differentiating both sides with respect to x,
`y d/dx log cos x + log cos x d/dx (y)= x d/dx log cos y + log cos y d/dx (x)`
⇒ `y * 1/(cos x) d/dx cos x + log cos x * dy/dx= x * 1/(cos y) d/dx cos y + log cos y xx 1`
⇒ `y * 1/(cos x) (- sin x) + log cos x. dy/dx = x 1/(cos y) (-sin y) dy/dx + log cos y`
⇒ `-y tan x + log cos x dy/dx = - x tan y dy/dx + log cos y`
⇒ `log cos x dy/dx + x tan y dy/dx = log cos y + y tan x`
⇒ `dy/dx (log cos x + x tan y) = log cos y + y tan x`
`therefore dy/dx = (log cos y + y tan x)/ (log cos x + x tan y)`
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