91Ó°ÊÓ

In inverse trigonometric functions, we reverse the roles of the trigonometric functions. For example, the inverse sine function, represented as \(\sin^{-1}(x)\) or \(\arcsin(x)\), gives the angle \(\theta\) whose sine is \(x\). In the given exercise, we encounter \(\sin^{-1}(x^4)\). This function returns an angle whose sine is \(x^4\). It's important to remember that the range for \(\arcsin(x)\) is between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\).

When differentiating inverse trigonometric functions, special rules apply. For example, the derivative of \(\sin^{-1}(u)\) with respect to \(u\) is \(\frac{1}{\sqrt{1-u^2}}\). Knowing these formulas is critical as they directly apply to the differentiation process using the chain rule.

This understanding helps us break down complex functions that involve inverse trigonometric components.

A composite function is formed when we plug one function into another, such as \( f(x) = 6 \sin^{-1}(x^4) \). In this case, \(\sin^{-1}(x^4)\) is a composite function where \(x^4\) is inside the inverse sine function.

To differentiate composite functions, we use the chain rule. The chain rule states that \( (f(g(x)))' = f'(g(x)) \, g'(x) \). This means we first find the derivative of the outer function and then multiply it by the derivative of the inner function.

For the given exercise:

• We identified our outer function as \(6 \sin^{-1}(u)\) with \(u = x^4\).
• We differentiated the outer function to get \( \frac{d}{du}[6 \sin^{-1}(u)] = 6 \cdot \frac{1}{\sqrt{1-u^2}} \).
• Then, we differentiated the inner function \( u = x^4 \) to get \( \frac{d}{dx}[x^4] = 4x^3 \).

Putting it all together using the chain rule, we get:\[ f'(x) = \left(6 \cdot \frac{1}{\sqrt{1-(x^4)^2}}\right) \cdot (4x^3) \].

Simplifying expressions is often necessary to make the math more manageable and to provide the final form of the answer. After applying the chain rule in our differentiation problem, we had an expression that required some simplification.

Let's break it down:

We started with:

• \( f'(x) = \frac{24x^3}{\sqrt{1-(x^4)^2}} \)

Here, \((x^4)^2\) simplifies to \(x^8\). So, we rewrite our expression:

• \( f'(x) = \frac{24x^3}{\sqrt{1-x^8}} \)

This final expression is straightforward and shows the full result of differentiating the original function \(6 \sin^{-1}(x^4)\).

By recognizing organized steps to simplify, we make sure the expressions are not only correct but also easy to interpret. Thus, we achieve clear and concise solutions, which are crucial for understanding complex calculus concepts.