To find the derivative of the function $y = (\cos x)^x + x^{\log x}$, we need to apply the rules of differentiation, including the chain rule, product rule, and logarithmic differentiation.

1. Differentiate $y = (\cos x)^x$:

Let $u(x) = (\cos x)^x$. To differentiate this, use logarithmic differentiation.

$\ln u(x) = \ln \left( (\cos x)^x \right)$ $\ln u(x) = x \ln(\cos x)$

Now, differentiate both sides with respect to $x$:

$\frac{1}{u(x)} \frac{du(x)}{dx} = \ln(\cos x) + x \cdot \frac{-\sin x}{\cos x}$ $\frac{du(x)}{dx} = u(x) \left( \ln(\cos x) - x \tan x \right)$ $\frac{du(x)}{dx} = (\cos x)^x \left( \ln(\cos x) - x \tan x \right)$

2. Differentiate $y = x^{\log x}$:

Let $v(x) = x^{\log x}$. Again, use logarithmic differentiation.

$\ln v(x) = \log x \cdot \ln x$

Differentiate both sides with respect to $x$:

$\frac{1}{v(x)} \frac{dv(x)}{dx} = \frac{1}{x} \ln x + \frac{\log x}{x}$ $\frac{dv(x)}{dx} = x^{\log x} \left( \frac{\ln x + \log x}{x} \right)$

3. Combine the derivatives:

Now, the derivative of the entire function $y$ is:

$\frac{dy}{dx} = \frac{du(x)}{dx} + \frac{dv(x)}{dx}$ $\frac{dy}{dx} = (\cos x)^x \left( \ln(\cos x) - x \tan x \right) + x^{\log x} \left( \frac{\ln x + \log x}{x} \right)$

This is the derivative of the function $y = (\cos x)^x + x^{\log x}$.

Would you like a step-by-step breakdown or further explanation on any of the differentiation techniques used here?

Here are some related questions:

1. How would you differentiate $y = (\sin x)^x$?
2. What is the derivative of $y = x^{\sin x}$?
3. Can you find the second derivative of $y = (\cos x)^x + x^{\log x}$?
4. How do you apply logarithmic differentiation to a function of the form $y = a^x$?
5. What is the derivative of $y = e^{x \ln x}$?

Tip: Logarithmic differentiation is a powerful technique when dealing with functions where the variable is both the base and the exponent.