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

The Correct Option is A

To solve for \(\frac{dy}{dx}\) given that \((\cos x)^y = (\sin y)^x\), we start by taking the natural logarithm on both sides of the equation: 

\(\ln((\cos x)^y) = \ln((\sin y)^x)\)

This gives:

\(y \ln(\cos x) = x \ln(\sin y)\)

Now, differentiate both sides with respect to \(x\). For the left side, apply the product rule:

\(\frac{d}{dx}(y \ln(\cos x)) = y \cdot \frac{d}{dx}(\ln(\cos x)) + \ln(\cos x) \cdot \frac{dy}{dx}\)

\(\frac{d}{dx}(\ln(\cos x)) = \frac{1}{\cos x} \cdot (-\sin x) = -\tan x\)

Thus, the derivative of the left side is:

\(y (-\tan x) + \ln(\cos x) \cdot \frac{dy}{dx}\)

For the right side, apply the product rule:

\(\frac{d}{dx}(x \ln(\sin y)) = x \cdot \frac{d}{dx}(\ln(\sin y)) + \ln(\sin y) \cdot \frac{d}{dx}(x)\)

\(\frac{d}{dx}(\ln(\sin y)) = \frac{1}{\sin y}\cdot\cos y\cdot \frac{dy}{dx} = \cot y \cdot \frac{dy}{dx}\)

The derivative of the right side is:

\(x \cdot \cot y \cdot \frac{dy}{dx} + \ln(\sin y)\)

Equate the derivatives:

\(y (-\tan x) + \ln(\cos x) \cdot \frac{dy}{dx} = x \cdot \cot y \cdot \frac{dy}{dx} + \ln(\sin y)\)

Rearrange to solve for \(\frac{dy}{dx}\):

\(\ln(\cos x) \cdot \frac{dy}{dx} - x \cdot \cot y \cdot \frac{dy}{dx} = \ln(\sin y) + y \tan x\)

Factor out \(\frac{dy}{dx}\):

\(\frac{dy}{dx}(\ln(\cos x) - x \cdot \cot y) = \ln(\sin y) + y \tan x\)

Solve for \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \frac{\ln(\sin y) + y \tan x}{\ln(\cos x) - x \cdot \cot y}\)

Thus, the solution is:

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

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:

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