Problem 5 - Technical Mathematics with Calculus

Question

Find the derivative. $$y=\sin ^{-1} \frac{\sin x-\cos x}{\sqrt{2}}$$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = \frac{1}{\sqrt{1-\left(\frac{\sin x - \cos x}{\sqrt{2}}\right)^2}} \cdot \frac{1}{\sqrt{2}}(\cos x + \sin x) \)

Step 1: Apply the chain rule for differentiation

The chain rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function. Here, let \( u = \frac{\sin x - \cos x}{\sqrt{2}} \). Hence, we need to find the derivative of \( \sin^{-1}(u) \) and multiply it by the derivative of \( u \) with respect to \( x \).

Step 2: Differentiate the outer function

The outer function is \( y = \sin^{-1}(u) \). The derivative of \( \sin^{-1}(u) \) with respect to \( u \) is \( \frac{1}{\sqrt{1-u^2}} \).

Step 3: Differentiate the inner function

Differentiate the inner function \( u \) with respect to \( x \), which gives \( u' = \frac{1}{\sqrt{2}}(\cos x + \sin x) \).

Step 4: Combine using the chain rule

Combine the results of Step 2 and Step 3 using the chain rule to find the overall derivative: \[ \frac{dy}{dx} = \frac{1}{\sqrt{1-u^2}} \cdot \frac{1}{\sqrt{2}}(\cos x + \sin x) \].

Step 5: Substitute back the value of \( u \)

Substitute back the value of \( u = \frac{\sin x - \cos x}{\sqrt{2}} \) into the derivative found in Step 4 to get the final answer: \[ \frac{dy}{dx} = \frac{1}{\sqrt{1-\left(\frac{\sin x - \cos x}{\sqrt{2}}\right)^2}} \cdot \frac{1}{\sqrt{2}}(\cos x + \sin x) \].

Key Concepts

Derivative of Inverse Trigonometric Functions

Understanding the derivatives of inverse trigonometric functions is crucial for comprehending various calculus problems. Inverse trigonometric functions like \( \sin^{-1}(x) \), \( \cos^{-1}(x) \), and \( \tan^{-1}(x) \) have derivatives that are expressed in terms of fractions involving the square root of \( 1 \) minus the square of the function's argument.

For instance, the derivative of \( \sin^{-1}(x)\) is \( \frac{1}{\sqrt{1-x^2}} \), which applies when \( x \) is within the domain \(-1 \leq x \leq 1\). Notably, this derivative shows the rate of change of the angle whose sine is \( x \) with respect to changes in \( x \).

Example:

If you're dealing with a function like \( y = \sin^{-1}(u) \), where \( u \) is a function of \( x \), differentiating with respect to \( x \) involves recognizing that \( y \) is a composite function. Then, you'll need to use the chain rule, which leads to \( \frac{dy}{dx} = \frac{1}{\sqrt{1-u^2}} \times \frac{du}{dx} \), indicating the close relationship between the derivatives of inverse trigonometric functions and the chain rule.

Composite Function Differentiation

Differentiation of composite functions is performed using the chain rule. A composite function is formed when one function is nested inside another, such as \( f(g(x)) \). To differentiate a composite function, you differentiate the outer function and multiply it by the derivative of the inner function.

Notable Point:

The outer function is effectively 'composed' with the inner function, and each part must be differentiated in turn. For example, if \( f(x) = \sin(x) \) and \( g(x) = x^2 \), and you have \( y = f(g(x)) \), then \( y = \sin(x^2) \), which is a composite function. The derivative is found by taking \( f'(g(x)) \) and then multiplying it by \( g'(x) \). Conceptually, this is like peeling the layers of an onion—each layer (function) must be addressed in the differentiation process.

Applying the Chain Rule

Applying the chain rule can seem daunting at first, but it's a powerful tool for finding derivatives of composite functions. When faced with a composite function like \( y = f(g(x)) \), the chain rule tells us to differentiate the outer function \( f \) at the inner function \( g \) and multiply by the derivative of \( g \) with respect to \( x \).

Practical Steps:

In practice, follow these steps to apply the chain rule:

• Identify the inner function \( g(x) \) and differentiate it to get \( g'(x) \).
• Identify the outer function \( f(u) \), with \( u = g(x) \), and differentiate it with respect to \( u \) to get \( f'(u) \).
• Multiply the derivatives \( f'(u) \) and \( g'(x) \) to find the total derivative \( \frac{dy}{dx} \).

In the given exercise, the use of the chain rule transforms the process into manageable steps, ensuring that the student can confidently find the derivative of complex functions by treating each part sequentially.