Find dy/dx, if (cos x)y = (cos y)x

Question

Q) Find $\frac{dy}{dx}$, if $(\cos x)^y = (\cos y)^x$

Sapling Academy · Accepted Answer

Given $(\cos x)^y = (\cos y)^x$
Taking log both sides

log $(\cos x)^y$ = log $(\cos y)^x$

Since log (a) b = b. log a

∴ y log (cos x) = x log (cos y)

Differentiating both sides w.r.t. x

$\frac{d}{dx}$ (y log cos x) = $\frac{d}{dx}$ (x log cos y)

$\frac{d (y)}{dx}$ . log cos x + $\frac{d (log \cos x)}{dx}$ . y  = $\frac{d (x)}{dx}$ . log cos y + $\frac{d (log \cos y)}{dx}$ . x

$\frac{dy}{dx}$ . log cos x + $\frac{1}{\cos x} . \frac{d (\cos x)}{dx}$ . y  = 1 . log cos y + $\frac{1}{\cos y} . \frac{d (\cos y)}{dx}$ . x

$\frac{dy}{dx}$ . log cos x + $\frac{1}{\cos x} $. (- sin x) . y  = log cos y + $\frac{1}{\cos y} . \frac{d (\cos y)}{dx} . \frac{dy}{dy}$. x

$\frac{dy}{dx}$ . log cos x - tan x . y  = log cos y + $\frac{1}{\cos y} . \frac{d (\cos y)}{dy} . \frac{dy}{dx}$. x

$\frac{dy}{dx}$ . log cos x - y tan x  = log cos y + x . $\frac{1}{\cos y}$ . (- sin y) . $\frac{dy}{dx}$

$\frac{dy}{dx}$ . log cos x - y tan x  = log cos y - x tan y . $\frac{dy}{dx}$

$\frac{dy}{dx}$ . log cos x + x tan y . $\frac{dy}{dx}$  = log cos y  + y tan x

$\frac{dy}{dx}$ ( log cos x + x tan y)  = log cos y  + y tan x

$\frac{dy}{dx} = \frac{(log \cos y  + y 	an x )}{ ( log \cos x + x 	an y) }$ .... Final Answer

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