91Ó°ÊÓ

When tackling the problem of finding the derivative of a function like the one in the exercise, the **Product Rule** plays a pivotal role. The Product Rule is a method in calculus used to differentiate products of two functions, which is essential for our problem where we have a product of an explicit function and an implicit function.Given two differentiable functions, say:

• \( f(x) \) and \( g(x) \)

The derivative of their product is:

• \((f \cdot g)' = f' \cdot g + f \cdot g' \)

To apply this to the function \( x \sqrt{4-x^2} \), you consider \( f(x) = x \) and \( g(x) = \sqrt{4-x^2}\). - The derivative of \( f(x) \) is straightforward: \( f'(x) = 1 \).- For \( g(x) \), the **Chain Rule** helps in finding \( g'(x) \).Utilizing the product rule, the derivative becomes:

• \( 1 \cdot \sqrt{4-x^2} + x \cdot g'(x) \)

In our solution, this gives:

• \[ \sqrt{4-x^2} - \frac{x^2}{\sqrt{4-x^2}} \] which, when calculated properly, smoothens the way forward for simplifying the derivative of intricate functions composed of products.

The **Chain Rule** is another cornerstone of calculus, used to find the derivative of a composite function. This rule aids in differentiating a function that contains another function inside it, which is particularly useful here, especially for the \( g(x) = \sqrt{4-x^2} \) part and \( 4 \arcsin \frac{x}{2} \) from the original function.The Chain Rule states that for a composite function \( y = f(g(x)) \), the derivative \( y' \) is:

• \( f'(g(x)) \cdot g'(x) \)

In our specific example:- For \( \sqrt{4-x^2} \), the outer function is the square root \( ()^{1/2} \), and the inner is \( 4-x^2 \). - Applying the chain rule gives: \( \frac{-x}{\sqrt{4-x^2}} \).- For \( \arcsin \frac{x}{2} \), the outer function is \( \arcsin \) and the inner is \( \frac{x}{2} \). - Upon applying the chain rule: \( \frac{2}{\sqrt{4-x^2}} \) It connects seamlessly with other rules, helping to break down complex derivatives into manageable pieces.

Calculating the derivative of an inverse trigonometric function, such as **Arcsin**, is integral for solving our exercise. Specifically, we need to find \( 4 \arcsin \frac{x}{2} \) in the given expression.The derivative of \( \arcsin u \) with respect to \( u \) is:

• \( \frac{1}{\sqrt{1-u^2}} \)

When the function inside arcsin is itself a function of \( x \), as in our function \( \frac{x}{2} \), the Chain Rule also comes into play:

• We take the derivative \( \frac{1}{2} \) of \( \frac{x}{2} \).

Combining these two steps, the derivative for our function part turns into:

• \( 4 \times \frac{1}{2} \times \frac{1}{\sqrt{1-(\frac{x}{2})^2}} \)

Simplified further using the identities of trigonometric derivatives:

• \( \frac{2}{\sqrt{4-x^2}} \)

This derivative captures how the inverse sine function changes and provides an internal perspective into solving complex derivative problems.