1. **Problem statement:** Differentiate the following functions: a) $g(x) = (x^2 - 2)(2x + 3)$ b) $f(x) = \frac{2x - 5}{x + 1}$ c) $y = (x^2 + 1)^7$ --- 2. **Formulas and rules:** - For product rule: $\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$ - For quotient rule: $\frac{d}{dx}\left[\frac{u(x)}{v(x)}\right] = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$ - For chain rule: $\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$ --- 3. **Solution for a):** Let $u = x^2 - 2$, $v = 2x + 3$ Calculate derivatives: $u' = 2x$ $v' = 2$ Apply product rule: $$g'(x) = u'v + uv' = (2x)(2x + 3) + (x^2 - 2)(2)$$ Expand: $$= 4x^2 + 6x + 2x^2 - 4 = 6x^2 + 6x - 4$$ --- 4. **Solution for b):** Let $u = 2x - 5$, $v = x + 1$ Calculate derivatives: $u' = 2$ $v' = 1$ Apply quotient rule: $$f'(x) = \frac{u'v - uv'}{v^2} = \frac{2(x + 1) - (2x - 5)(1)}{(x + 1)^2}$$ Simplify numerator: $$= \frac{2x + 2 - 2x + 5}{(x + 1)^2} = \frac{7}{(x + 1)^2}$$ --- 5. **Solution for c):** Let $u = x^2 + 1$ Then $y = u^7$ Calculate derivative of $u$: $u' = 2x$ Apply chain rule: $$\frac{dy}{dx} = 7u^{6} \cdot u' = 7(x^2 + 1)^6 (2x) = 14x (x^2 + 1)^6$$ --- **Final answers:** a) $g'(x) = 6x^2 + 6x - 4$ b) $f'(x) = \frac{7}{(x + 1)^2}$ c) $\frac{dy}{dx} = 14x (x^2 + 1)^6$