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When differentiating a function that is the product of two other functions, the product rule is your go-to method. Imagine you have a function like \( f(x) = u(x) \, v(x) \). To find its derivative, employ the formula:

• \( f'(x) = u'(x) \, v(x) + u(x) \, v'(x) \)

This means you differentiate the first function and multiply it by the second, then add the product of the first function and the derivative of the second.
In the given exercise, the term \( x \sin^{-1}(x) \) is differentiated using this rule. Here, \( u = x \) and \( v = \sin^{-1}(x) \). Applying the product rule gives us the derivative as \( \sin^{-1}(x) + \frac{x}{\sqrt{1-x^2}} \). This clear structure makes the product rule a favorite in calculus, especially when dealing with functions that naturally multiply together.

The chain rule is essential when dealing with composite functions, meaning a function nested within another function. If you have a function \( f(g(x)) \), the chain rule tells you how to find the derivative:

• Differentiate the outer function\( f \) with respect to the inner function \( g \).
• Multiply by the derivative of the inner function \( g(x) \).

This two-step approach lets you peel back layers of complexity in calculus problems.
In our exercise, we used the chain rule to find the derivative of \( \sqrt{1-x^2} \). The outer function in this case is \( \sqrt{u} \) and the inner function is \( 1-x^2 \). Following the chain rule, the derivative turns into \( \frac{-x}{\sqrt{1-x^2}} \). Seeing how each part fits together is key to mastering the chain rule.

Inverse trigonometric functions, like \( \sin^{-1}(x) \) which appears in our exercise, have derivatives that are just as crucial to know as the functions themselves. When differentiating such functions, remember:

• The derivative of \( \sin^{-1}(x) \) is \( \frac{1}{\sqrt{1-x^2}} \).
• This result arises because these functions undo the action of their corresponding trigonometric functions.

Using these derivatives in problems helps untangle complex expressions involving arcsines, arccosines, and other inverse trigonometric functions.
In our solution, the inverse sine function was differentiated, making \( \sin^{-1}(x) \) a part of the final result. Understanding how to use these derivatives allows you to tackle equations involving inverse trigonometric forms with ease.