Let's analyze and solve the problem step by step.

(b) Show that $\frac{dy}{dx} = \frac{20}{(x+1)^3}$:

Given: $y = \frac{3x^2 + 6x - 7}{(x+1)^2}$ We need to find $\frac{dy}{dx}$.

To differentiate $y$ with respect to $x$, we'll use the quotient rule. The quotient rule states: $\frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{(v(x))^2}$ Here, $u(x) = 3x^2 + 6x - 7$ and $v(x) = (x+1)^2$.

First, differentiate $u(x)$ and $v(x)$:

• $u'(x) = \frac{d}{dx}(3x^2 + 6x - 7) = 6x + 6$
• $v(x) = (x+1)^2$, so $v'(x) = 2(x+1)$

Now, apply the quotient rule: $\frac{dy}{dx} = \frac{(x+1)^2 \cdot (6x + 6) - (3x^2 + 6x - 7) \cdot 2(x+1)}{(x+1)^4}$ Simplify the numerator: $= \frac{6(x+1)(x+1)^2 - 2(3x^2 + 6x - 7)(x+1)}{(x+1)^4}$ Factor out $(x+1)$ from the numerator: $= \frac{(x+1)[6(x+1)^2 - 2(3x^2 + 6x - 7)]}{(x+1)^4}$ Cancel out $(x+1)$ from the numerator and denominator: $= \frac{6(x+1)^2 - 2(3x^2 + 6x - 7)}{(x+1)^3}$ Now simplify the expression further: $= \frac{6(x^2 + 2x + 1) - (6x^2 + 12x - 14)}{(x+1)^3}$ Expand and combine like terms: