Question

If $(\cos x)^y = (\cos y)^x$, then $\frac{dy}{dx}$ is:

• A. $\frac{ y \tan x + \log (\cos y) }{ x \tan y - \log (\cos x) }$
• B. $\frac{ x \tan y + \log (\cos x) }{ y \tan x + \log (\cos y) }$
• C. $\frac{ y \tan x - \log (\cos y) }{ x \tan y - \log (\cos x) }$
• D. $\frac{ y \tan x + \log (\cos y) }{ x \tan y + \log (\cos x) }$

Hint:

Use log to simplify exponents before differentiating.

Answer 1

The Correct Option is A

Take log: \[ \ln [ (\cos x)^y ] = \ln [ (\cos y)^x ]. \implies y \ln (\cos x) = x \ln (\cos y). \] Differentiate w.r.t. $x$: \[ y (-\tan x) + \ln (\cos x) \frac{dy}{dx} = \ln (\cos y) - x \tan y \frac{dy}{dx}. \] Collect: \[ \ln (\cos x) \frac{dy}{dx} + x \tan y \frac{dy}{dx} = y \tan x + \ln (\cos y). \] So, \[ \frac{dy}{dx} = \frac{ y \tan x + \ln (\cos y) }{ x \tan y - \ln (\cos x) }. \]