Question

Compute the derivative of the given function in two ways: (a) By simplifying first, then taking the derivative, and (b) by using the Chain Rule first then simplifying. Verify that the two answers are the same. $$f(x)=\sin \left(\sin ^{-1} x\right)$$

Short answer

The derivative of the function is 1 using both methods.

Step-by-step solution

Simplify the Function

First, realize that \( \sin(\sin^{-1}(x)) = x \) because \( \sin^{-1}(x) \) is the inverse of \( \sin(x) \). Thus, the function simplifies to \( f(x) = x \).

Find the Derivative of the Simplified Function

The derivative of \( f(x) = x \) with respect to \( x \) is \( f'(x) = 1 \).

Apply the Chain Rule

For the original function \( f(x) = \sin(\sin^{-1}(x)) \), let \( g(x) = \sin^{-1}(x) \) and \( h(x) = \sin(x) \). Hence, \( f(x) = h(g(x)) \). Using the chain rule, \( f'(x) = h'(g(x)) \cdot g'(x) \).

Calculate Each Derivative Using the Chain Rule

The derivative of \( h(x) = \sin(x) \) is \( h'(x) = \cos(x) \). Substitute \( x \) with \( g(x) = \sin^{-1}(x) \), so \( h'(g(x)) = \cos(\sin^{-1}(x)) \). The derivative of \( g(x) = \sin^{-1}(x) \) is \( g'(x) = \frac{1}{\sqrt{1-x^2}} \).

Simplify Using Trigonometric Identity

Now simplify \( h'(g(x)) = \cos(\sin^{-1}(x)) \). We know from the identity \( \cos(\sin^{-1}(x)) = \sqrt{1-x^2} \). So, \( h'(g(x)) = \sqrt{1-x^2} \).

Combine Results Using Chain Rule

Using the results from the previous steps, the derivative is \( f'(x) = \sqrt{1-x^2} \cdot \frac{1}{\sqrt{1-x^2}} = 1 \).

Verify Consistency

Both the simplification method and the chain rule method yield the same derivative: \( f'(x) = 1 \). This verifies that our calculations are consistent.

Key concepts

Chain Rule

The chain rule is an important concept in calculus when dealing with composite functions. A composite function is a function that is nested within another function, like in our original exercise where we have \( f(x) = \sin(\sin^{-1}(x)) \). When you have a function composed of other functions, the chain rule helps you find the derivative effectively. The rule states that the derivative of a composite function \( f(g(x)) \) is the derivative of the outer function \( h(x) \) evaluated at \( g(x) \) multiplied by the derivative of the inner function \( g(x) \). This can be written as:

• \( f'(x) = h'(g(x)) \cdot g'(x) \)

Applying this to our exercise, with \( g(x) = \sin^{-1}(x) \) and \( h(x) = \sin(x) \), we calculate each part. The outer function \( h(x) = \sin(x) \) has a derivative \( h'(x) = \cos(x) \), and because of the composition, you replace \( x \) with \( g(x) \), leading to \( h'(g(x)) = \cos(\sin^{-1}(x)) \). The derivative of the inner function \( g(x) = \sin^{-1}(x) \) gives \( g'(x) = \frac{1}{\sqrt{1-x^2}} \). Combining these as per the chain rule results in us calculating the derivative effectively.

Inverse Trigonometric Functions

Inverse trigonometric functions are the inverses of the basic trigonometric functions (like sine, cosine, and tangent). For example, \( \sin^{-1}(x) \) is the inverse of the \( \sin(x) \) function. These functions are crucial in calculus because they allow us to solve equations involving trigonometric identities and apply derivatives to them.

When dealing with the derivative of inverse trigonometric functions, specific formulas can be used. In our problem, we're using \( \sin^{-1}(x) \), whose derivative is:

• \( \frac{d}{dx} \sin^{-1}(x) = \frac{1}{\sqrt{1-x^2}} \)

This formula stems from the fact that inverse functions undo the action of their original functions. It is very useful when combined with the chain rule for dealing with composite functions like \( \sin(\sin^{-1}(x)) \), as it aids in transitioning back to simpler identities.

Simplification in Calculus

Simplification in calculus involves reducing complex expressions into simpler forms that are easier to work with. When tackling the problem of finding a derivative, it sometimes helps to simplify the function first before calculating the derivative.

Our exercise demonstrates this method effectively. The function \( f(x) = \sin(\sin^{-1}(x)) \) simplifies directly to \( f(x) = x \), using the fact that \( \sin(\sin^{-1}(x)) = x \) due to the inverse relationship between sine and arcsine.

Once simplified, the derivative calculation becomes straightforward:

• The derivative of \( f(x) = x \) is simply \( f'(x) = 1 \).

This simplification not only makes the problem easier but also serves as a verification method when compared with calculations using the chain rule. Simplification helps reveal underlying identities and relationships within a function that may not be immediately obvious, aiding in solving calculus problems with more confidence and less error.