You are asked to use logarithmic differentiation to find the derivative of the function:

$y = (\cos x)^x$

Steps for Logarithmic Differentiation:

1. Take the natural logarithm of both sides to make the exponent easier to handle: $\ln y = \ln \left( (\cos x)^x \right)$ Using the property of logarithms, $\ln(a^b) = b \ln(a)$, we simplify the right-hand side: $\ln y = x \ln(\cos x)$

2. Differentiate both sides with respect to $x$:

   • On the left-hand side, using the chain rule: $\frac{d}{dx} \ln y = \frac{1}{y} \frac{dy}{dx}$
   • On the right-hand side, apply the product rule to $x \ln(\cos x)$: $\frac{d}{dx} \left( x \ln(\cos x) \right) = \ln(\cos x) \cdot \frac{d}{dx}(x) + x \cdot \frac{d}{dx}(\ln(\cos x))$ The derivative of $x$ is 1, and the derivative of $\ln(\cos x)$ is: $\frac{d}{dx} \ln(\cos x) = \frac{-\sin x}{\cos x} = -\tan x$ Therefore, the derivative of $x \ln(\cos x)$ is: $\ln(\cos x) - x \tan x$

3. Combine the results: $\frac{1}{y} \frac{dy}{dx} = \ln(\cos x) - x \tan x$

4. Solve for $\frac{dy}{dx}$ by multiplying both sides by $y$: $\frac{dy}{dx} = y \left( \ln(\cos x) - x \tan x \right)$ Since $y = (\cos x)^x$, substitute it back: $\frac{dy}{dx} = (\cos x)^x \left( \ln(\cos x) - x \tan x \right)$

This is the derivative of the function $y = (\cos x)^x$ using logarithmic differentiation.

Would you like more details or have any questions? Here are five related questions you might find interesting:

1. How does the chain rule apply in logarithmic differentiation?
2. Why do we need logarithmic differentiation for functions like $(\cos x)^x$?
3. Can logarithmic differentiation be used for functions with other forms, like $x^x$?
4. How does the product rule work in this context of differentiating $x \ln(\cos x)$?
5. What are other common situations where logarithmic differentiation simplifies the problem?

Tip:

Logarithmic differentiation is particularly useful for functions that involve variables in both the base and the exponent, simplifying otherwise difficult derivatives.