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Given the equation $(cos~x)^{y}=(cos~y)^{x}$, we take the natural logarithm of both sides to simplify the exponents. $ln((cos~x)^{y}) = ln((cos~y)^{x})$ Using the power rule of logarithms, we get: $y~ln(cos~x) = x~ln(cos~y)$
Now, we differentiate both sides of the equation $y~ln(cos~x) = x~ln(cos~y)$ with respect to $x$ using the product rule and chain rule. $\frac{d}{dx}[y~ln(cos~x)] = \frac{d}{dx}[x~ln(cos~y)]$ Applying the product rule on both sides: $\frac{dy}{dx}ln(cos~x) + y\frac{d}{dx}ln(cos~x) = ln(cos~y) + x\frac{d}{dx}ln(cos~y)$
Now, we apply the chain rule to differentiate $ln(cos~x)$ and $ln(cos~y)$ with respect to $x$. $\frac{d}{dx}ln(cos~x) = \frac{1}{cos~x} * (-sin~x) = -tan~x$ $\frac{d}{dx}ln(cos~y) = \frac{1}{cos~y} * (-sin~y) * \frac{dy}{dx} = -tan~y * \frac{dy}{dx}$ Substituting these derivatives back into the equation: $\frac{dy}{dx}ln(cos~x) + y(-tan~x) = ln(cos~y) + x(-tan~y)\frac{dy}{dx}$ $\frac{dy}{dx}ln(cos~x) - y~tan~x = ln(cos~y) - x~tan~y~\frac{dy}{dx}$
Now, we rearrange the equation to isolate $\frac{dy}{dx}$ terms on one side: $\frac{dy}{dx}ln(cos~x) + x~tan~y~\frac{dy}{dx} = ln(cos~y) + y~tan~x$ Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}[ln(cos~x) + x~tan~y] = ln(cos~y) + y~tan~x$ Finally, solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{ln(cos~y) + y~tan~x}{ln(cos~x) + x~tan~y}$
Final Answer: $\frac{dy}{dx} = \frac{ln(cos~y) + y~tan~x}{ln(cos~x) + x~tan~y}$
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