Question:
If $(\cos x)^y = (\cos y)^x$, then $\frac{dy}{dx}$ is:
If $(\cos x)^y = (\cos y)^x$, then $\frac{dy}{dx}$ is:
Show Hint
Use log to simplify exponents before differentiating.
Updated On: Jul 1, 2025
- $\frac{ y \tan x + \log (\cos y) }{ x \tan y - \log (\cos x) }$
- $\frac{ x \tan y + \log (\cos x) }{ y \tan x + \log (\cos y) }$
- $\frac{ y \tan x - \log (\cos y) }{ x \tan y - \log (\cos x) }$
- $\frac{ y \tan x + \log (\cos y) }{ x \tan y + \log (\cos x) }$
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The Correct Option is A
Solution and Explanation
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,
\[
\frac{dy}{dx} = \frac{ y \tan x + \ln (\cos y) }{ x \tan y - \ln (\cos x) }.
\]
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