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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:
\[
= \frac{6x^2 + 12x + 6 - 6x^2 - 12x + 14}{(x+1)^3}
= \frac{20}{(x+1)^3}
\]
This matches the required expression, so the result is proven.

### (c) Find \(\frac{d^2y}{dx^2}\) and the real values of \(x\) for which \(\frac{d^2y}{dx^2} = \frac{15}{4}\):

Now, let's find the second derivative \(\frac{d^2y}{dx^2}\) from the result of \(\frac{dy}{dx}\).

Given:
\[
\frac{dy}{dx} = \frac{20}{(x+1)^3}
\]
Differentiate \(\frac{dy}{dx}\) again with respect to \(x\):
\[
\frac{d^2y}{dx^2} = \frac{d}{dx} \left(\frac{20}{(x+1)^3}\right)
\]
This can be differentiated using the chain rule:
\[
\frac{d^2y}{dx^2} = 20 \cdot \frac{d}{dx} \left((x+1)^{-3}\right)
= 20 \cdot (-3)(x+1)^{-4} \cdot 1
= \frac{-60}{(x+1)^4}
\]
Now, solve for \(x\) when:
\[
\frac{-60}{(x+1)^4} = \frac{15}{4}
\]
Cross-multiply to solve for \((x+1)^4\):
\[
-60 \cdot 4 = 15 \cdot (x+1)^4
\]
\[
-240 = 15(x+1)^4
\]
\[
(x+1)^4 = -\frac{240}{15} = -16
\]
Since a fourth power cannot be negative, this equation has no real solution. Hence, there are no real values of \(x\) for which \(\frac{d^2y}{dx^2} = \frac{15}{4}\).

Would you like more details on any step or further questions?

Here are 5 related questions:
1. What is the quotient rule in differentiation, and when should it be used?
2. How does the chain rule apply in differentiating functions of the form \(f(g(x))\)?
3. Can you explain the significance of the second derivative in calculus?
4. How do you interpret an equation having no real solution in a real-world context?
5. What are the common methods to simplify complex algebraic expressions?

**Tip:** Always recheck your simplifications step by step to avoid algebraic errors, especially when dealing with polynomial functions and their derivatives.

Given that y = (3x^2 + 6x - 7)/(x+1)^2, x ≠ -1, show that dy/dx = 20/(x+1)^3. Hence find d^2y/dx^2 and the real values of x for which d^2y/dx^2 = 15/4.

This problem involves differentiating a rational function using the quotient rule and then finding the second derivative using the chain rule. Suitable for students in Grades 11-12, this detailed solution also explores why some equations may have no real solutions.

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