If \((\cos x)^y = (\sin y)^x\) then \(\displaystyle\frac{dy}{dx}\) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{\log_e\sin y + y\tan x}{\log_e\cos x - x\cot y}\)
Both sides have variable powers, so take logarithms:
\(y\log_e\cos x = x\log_e\sin y\)
Differentiating with respect to x, using the product rule on each side:
\( \displaystyle \frac{dy}{dx}\log_e\cos x + y\times\frac{-\sin x}{\cos x} = \log_e\sin y + x\times\frac{\cos y}{\sin y}\,\frac{dy}{dx} \)
\( \displaystyle \frac{dy}{dx}\log_e\cos x - y\tan x = \log_e\sin y + x\cot y\,\frac{dy}{dx} \)
Collecting the \(\frac{dy}{dx}\) terms:
\( \displaystyle \frac{dy}{dx}\left(\log_e\cos x - x\cot y\right) = \log_e\sin y + y\tan x \)
\( \displaystyle \frac{dy}{dx} = \frac{\log_e\sin y + y\tan x}{\log_e\cos x - x\cot y} \)