Let's solve the derivatives of the given functions using the chain rule.

Problem 13:

$y = \frac{2(x + 4)}{(x + 4)^2 - 1}$

We need to apply the quotient rule along with the chain rule here. Let's differentiate step by step.

Let:

• $u(x) = 2(x + 4)$
• $v(x) = (x + 4)^2 - 1$

The derivative of the quotient $\frac{u(x)}{v(x)}$ is given by the quotient rule:

$y' = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$

Steps:

1. Differentiate $u(x) = 2(x + 4)$: $u'(x) = 2$

2. Differentiate $v(x) = (x + 4)^2 - 1$ using the chain rule: $v'(x) = 2(x + 4)$

Now, apply the quotient rule:

[ y' = \frac{(2)[(x + 4)^2 - 1] - [2(x + 4)][2(x + 4)]}{[(x + 4)^2 - 1]^2} ]

This expression can be simplified further.

Problem 14:

$y = \frac{1}{(1 + \sqrt{1 - x})^2}$

For this function, we'll again apply the chain rule. We can let:

$y = (1 + \sqrt{1 - x})^{-2}$

We need to differentiate using the chain rule:

1. Start by differentiating the outer function: $y' = -2(1 + \sqrt{1 - x})^{-3}$

2. Now differentiate the inner function $1 + \sqrt{1 - x}$: $\frac{d}{dx}(1 + \sqrt{1 - x}) = \frac{-1}{2\sqrt{1 - x}}$

Finally, multiply these derivatives to get the complete derivative:

$y' = -2(1 + \sqrt{1 - x})^{-3} \cdot \frac{-1}{2\sqrt{1 - x}}$

This can also be simplified further.

Would you like me to proceed with simplifying the expressions or provide any more details? Here are some related questions to consider:

1. How do we apply the quotient rule for more complex functions?
2. What are some common mistakes when applying the chain rule?
3. How does the chain rule work in multi-variable calculus?
4. What are the conditions under which the chain rule is applicable?
5. How does the chain rule differ from other differentiation techniques?

Tip: When differentiating compositions of functions, always break down each component and apply the derivative rules step by step.