Question

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

• A. $\frac{\log_e \sin y - y \tan x}{\log_e \cos x + x \cot y}$
• B. $\frac{\log_e \sin y + y \tan x}{\log_e \cos x + x \cos y}$
• C. $\frac{\log_e \sin y + y \tan x}{\log_e \cos x - x \cot y}$
• D. $\frac{\log_e \cos x - x \cos y}{\log_e \sin y + y \tan x}$

Answer 1

Answer: (A)

$\frac{\log_e \sin y - y \tan x}{\log_e \cos x + x \cot y}$

Answer 2

We are given:

$(\cos x)^y = (\sin y)^x.$

Take the natural logarithm on both sides:

$y \log_e(\cos x) = x \log_e(\sin y).$

Differentiate both sides with respect to $x$ using implicit differentiation:

$\frac{d}{dx} \left[ y \log_e(\cos x) \right] = \frac{d}{dx} \left[ x \log_e(\sin y) \right].$

Using the product rule on both sides:

$\frac{dy}{dx} \log_e(\cos x) + y \times \frac{d}{dx} \left[ \log_e(\cos x) \right] = \log_e(\sin y) + x \times \frac{d}{dx} \left[ \log_e(\sin y) \right].$

Simplify each term:

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

Reorganize terms to isolate $\frac{dy}{dx}$:

$\frac{dy}{dx} \log_e(\cos x) - x \times \frac{\cos y}{\sin y} \times \frac{dy}{dx} = \log_e(\sin y) - y \tan x.$

Factor out $\frac{dy}{dx}$:

$\frac{dy}{dx} \left[ \log_e(\cos x) + x \cot y \right] = \log_e(\sin y) - y \tan x.$

Solve for $\frac{dy}{dx}$:

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

Thus, the correct answer is:

$\boxed{\frac{\log_e \sin y - y \tan x}{\log_e \cos x + x \cot y}}.$