Let's solve these derivative problems step by step.

Problem 11

The given function is:

$g(x) = \int_{5x}^{7x} \frac{u^2 - 1}{u^2 + 1} \, du$

We are asked to find $g'(x)$. This is a case of differentiating a definite integral with variable limits, so we will apply the Leibniz rule for differentiation under the integral sign. The rule is:

$\frac{d}{dx} \left( \int_{a(x)}^{b(x)} f(u) \, du \right) = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x)$

In our case:

• $f(u) = \frac{u^2 - 1}{u^2 + 1}$
• $a(x) = 5x$ and $b(x) = 7x$

Using the formula: $g'(x) = \frac{(7x)^2 - 1}{(7x)^2 + 1} \cdot 7 - \frac{(5x)^2 - 1}{(5x)^2 + 1} \cdot 5$

Simplifying both terms:

$g'(x) = \frac{49x^2 - 1}{49x^2 + 1} \cdot 7 - \frac{25x^2 - 1}{25x^2 + 1} \cdot 5$

Now, combining terms:

$g'(x) = \frac{7(49x^2 - 1)}{49x^2 + 1} - \frac{5(25x^2 - 1)}{25x^2 + 1}$

This is the expression for the derivative.

Problem 12

The function is:

$y = \int_{\cos(x)}^{\sin(x)} \ln(6 + 8v) \, dv$

Again, applying the Leibniz rule:

$y'(x) = \ln(6 + 8\sin(x)) \cdot \cos(x) - \ln(6 + 8\cos(x)) \cdot (-\sin(x))$

Simplifying the second term:

$y'(x) = \ln(6 + 8\sin(x)) \cdot \cos(x) + \ln(6 + 8\cos(x)) \cdot \sin(x)$

Thus, this is the expression for $y'(x)$.

Let me know if you'd like more details or clarifications on any steps!

Here are some related questions for deeper understanding:

1. How does Leibniz's rule generalize for cases with both upper and lower variable limits?
2. What is the intuition behind differentiating integrals with variable limits?
3. How does changing the integrand function affect the derivative?
4. Why is the chain rule necessary when using Leibniz's rule in these examples?
5. How would the result change if the limits were constants instead of functions of $x$?

Tip: For integrals with variable limits, always remember to apply the chain rule to the upper and lower limit functions as part of the differentiation process.