1. **State the problem:** We want to determine if the function $$f(x) = \frac{5\cos x}{2\tan x - 2}$$ is differentiable using the definition of the derivative: $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$ 2. **Recall the definition and conditions:** A function is differentiable at a point if this limit exists and is finite. 3. **Analyze the function:** The function is a quotient of trigonometric functions. We must check where the denominator is zero because the function is not defined there, and hence not differentiable. 4. **Find where denominator is zero:** $$2\tan x - 2 = 0 \implies \tan x = 1$$ This happens at $$x = \frac{\pi}{4} + k\pi, k \in \mathbb{Z}$$ At these points, $$f(x)$$ is not defined, so it is not differentiable there. 5. **Elsewhere, differentiability:** Since $$\cos x$$ and $$\tan x$$ are differentiable where defined, and the denominator is not zero, the quotient rule applies and $$f(x)$$ is differentiable. 6. **Conclusion:** - $$f(x)$$ is differentiable for all $$x$$ where $$2\tan x - 2 \neq 0$$. - $$f(x)$$ is not differentiable at points where $$\tan x = 1$$. Hence, $$f(x)$$ is differentiable except at $$x = \frac{\pi}{4} + k\pi$$ for integers $$k$$.