Question: diffrenciate $g(x)= e^x \cdot (e^x -2)^{-2}$
1. **State the problem:** Differentiate the function $$g(x) = e^x \cdot (e^x - 2)^{-2}$$ with respect to $$x$$.
2. **Recall the product rule:** If $$g(x) = u(x) \cdot v(x)$$, then $$g'(x) = u'(x) v(x) + u(x) v'(x)$$.
3. **Identify components:** Let $$u(x) = e^x$$ and $$v(x) = (e^x - 2)^{-2}$$.
4. **Differentiate $$u(x)$$:** $$u'(x) = \frac{d}{dx} e^x = e^x$$.
5. **Differentiate $$v(x)$$ using the chain rule:**
$$v(x) = (e^x - 2)^{-2}$$
Let $$w = e^x - 2$$, so $$v = w^{-2}$$.
Then $$v'(x) = -2 w^{-3} \cdot \frac{d}{dx} w = -2 (e^x - 2)^{-3} \cdot e^x$$.
6. **Apply the product rule:**
$$g'(x) = u'(x) v(x) + u(x) v'(x) = e^x (e^x - 2)^{-2} + e^x \left(-2 (e^x - 2)^{-3} e^x \right)$$
7. **Simplify:**
$$g'(x) = e^x (e^x - 2)^{-2} - 2 e^{2x} (e^x - 2)^{-3}$$
8. **Factor common terms:**
$$g'(x) = e^x (e^x - 2)^{-3} \left( (e^x - 2) - 2 e^x \right)$$
9. **Simplify inside the parentheses:**
$$(e^x - 2) - 2 e^x = e^x - 2 - 2 e^x = - e^x - 2$$
10. **Final derivative:**
$$g'(x) = e^x (e^x - 2)^{-3} (- e^x - 2) = - e^x (e^x + 2) (e^x - 2)^{-3}$$
**Answer:**
$$\boxed{g'(x) = - \frac{e^x (e^x + 2)}{(e^x - 2)^3}}$$