Differentiate y = (ln x)^ x + (sin x)^ x using logarithmic differentiation.

Take the natural logarithm of both sides:

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

Using the logarithm property $\ln(a^b) = b \ln a$:

$$\ln u = \cos x \cdot \ln(\ln x)$$

Now, differentiate both sides with respect to $x$ using the product rule on the right side:

$$\frac{1}{u} \frac{du}{dx} = \frac{d}{dx}(\cos x) \cdot \ln(\ln x) + \cos x \cdot \frac{d}{dx}(\ln(\ln x))$$

We know $\frac{d}{dx}(\cos x) = -\sin x$.
For $\frac{d}{dx}(\ln(\ln x))$, use the chain rule: $\frac{d}{dx}(\ln f(x)) = \frac{f'(x)}{f(x)}$. Here $f(x) = \ln x$, so $f'(x) = \frac{1}{x}$.

$$\frac{d}{dx}(\ln(\ln x)) = \frac{1}{\ln x} \cdot \frac{1}{x} = \frac{1}{x \ln x}$$

Substitute these derivatives back:

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

Multiply by $u$ to solve for $\frac{du}{dx}$:

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

Substitute $u = (\ln x)^{\cos x}$ back:

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

Take the natural logarithm of both sides:

$$\ln v = \ln \left((\sin x)^{\ln x}\right)$$

Using the logarithm property $\ln(a^b) = b \ln a$:

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

Now, differentiate both sides with respect to $x$ using the product rule on the right side:

$$\frac{1}{v} \frac{dv}{dx} = \frac{d}{dx}(\ln x) \cdot \ln(\sin x) + \ln x \cdot \frac{d}{dx}(\ln(\sin x))$$

We know $\frac{d}{dx}(\ln x) = \frac{1}{x}$.
For $\frac{d}{dx}(\ln(\sin x))$, use the chain rule:

$$\frac{d}{dx}(\ln(\sin x)) = \frac{1}{\sin x} \cdot \frac{d}{dx}(\sin x) = \frac{1}{\sin x} \cdot \cos x = \cot x$$

Substitute these derivatives back:

$$\frac{1}{v} \frac{dv}{dx} = \frac{1}{x} \cdot \ln(\sin x) + \ln x \cdot \cot x$$

Multiply by $v$ to solve for $\frac{dv}{dx}$:

$$\frac{dv}{dx} = v \left(\frac{\ln(\sin x)}{x} + \ln x \cdot \cot x\right)$$

Substitute $v = (\sin x)^{\ln x}$ back:

$$\frac{dv}{dx} = (\sin x)^{\ln x} \left(\frac{\ln(\sin x)}{x} + \ln x \cdot \cot x\right)$$

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

The final answer is $\boxed{(\ln x)^{\cos x} \left(-\sin x \cdot \ln(\ln x) + \frac{\cos x}{x \ln x}\right) + (\sin x)^{\ln x} \left(\frac{\ln(\sin x)}{x} + \ln x \cdot \cot x\right)}$.