Problem 3 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Differentiate the function given. \(y=\sin x \cdot \arcsin x\)

Step-by-Step Solution

Answer

The derivative of the function \(y = \sin x \cdot \arcsin x\) is \(y' = \cos x \cdot \arcsin x + \sin x \cdot \frac{1}{\sqrt{1 - x^2}}\).

Step 1: Identify the Functions in the Product

The function \(y = \sin x \cdot \arcsin x\) is a product of two functions, \(u = \sin x\) and \(v = \arcsin x\).

Step 2: Apply the Product Rule

According to the product rule, the derivative of \(uv\) is given by \(u'v+uv'\). Here, \(u = \sin x\), so \(u' = \cos x\), and \(v = \arcsin x\), so \(v' = \frac{1}{\sqrt{1 - x^2}}\).

Step 3: Substitute the Derivatives

Substitute \(u', u, v', v\) into the product rule formula to get \(y' = \cos x \cdot \arcsin x + \sin x \cdot \frac{1}{\sqrt{1 - x^2}}\).

Key Concepts

Differentiate Trigonometric Functions

In calculus, differentiating trigonometric functions is an essential skill that students must learn. This process involves finding the derivative of functions like sine, cosine, tangent, and their inverses. For example, the derivatives of the primary trigonometric functions include:

• \textbf{Sine (sin)}: The derivative of \(\sin x\) is \(\cos x\).
• \textbf{Cosine (cos)}: The derivative of \(\cos x\) is \(\sin x\).
• \textbf{Tangent (tan)}: The derivative of \(\tan x\) is \(\sec^2 x\).

Understanding these basic derivatives allows you to tackle more complex functions that involve trigonometry. As you practice, remember to also consider the chain rule when differentiating composite functions like \(\sin(g(x))\), where \(g(x)\) is another function.

Arcsin x Derivative

The inverse sine function, often written as \(\arcsin x\) or \(\sin^{-1} x\), is one of the inverses of the sine function. Differentiating \(\arcsin x\) is slightly more complex than differentiating standard trigonometric functions. The derivative of \(\arcsin x\) is given by: \[\frac{d}{dx}(\arcsin x) = \frac{1}{\sqrt{1 - x^2}}\] provided that \(x\) is in the domain of \(\arcsin\), which is \( -1 \leq x \leq 1 \). This result stems from the definition of the inverse sine and its relationship with the unit circle. It's important when applying this derivative to check that your function's domain is restricted such that \(\arcsin x\) is defined.

Applying The Product Rule

The product rule is a powerful differentiation rule used when dealing with the product of two functions. If you have a function \(y = u(x) \cdot v(x)\), the product rule states that the derivative of \(y\) with respect to \(x\) is: \[y' = u'v + uv'\]where \(u'\) is the derivative of \(u\) with respect to \(x\), and \(v'\) is the derivative of \(v\) with respect to \(x\). To apply the product rule effectively, follow these steps:

• Identify the two functions that are being multiplied (\(u\) and \(v\)).
• Find the derivative of each function individually (\(u'\) and \(v'\)).
• Apply the product rule formula, \(u'v + uv'\), to find the derivative of the product.

This rule is crucial when differentiating products of functions that cannot be simplified easily, such as the product of trigonometric functions (like \(\sin x\)) and their inverses (like \(\arcsin x\)), which cannot be combined into a single simple expression before differentiation.