Problem 5 - Thomas Calculus

Question

In Exercises \(1-8,\) given \(y=f(u)\) and \(u=g(x),\) find \(d y / d x=\) \(d y / d x=f^{\prime}(g(x)) g^{\prime}(x)\) $$ y=\sqrt{u}, \quad u=\sin x $$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = \frac{\cos x}{2 \sqrt{\sin x}} \)

Step 1: Identify the Functions

First, identify the given functions. We have two functions: \( y = \sqrt{u} \) and \( u = \sin x \).

Step 2: Find the Derivative of y with respect to u

Determine the derivative of \( y \) with respect to \( u \), denoted \( \frac{dy}{du} \). Since \( y = \sqrt{u} = u^{1/2} \), use the power rule:\[\frac{dy}{du} = \frac{1}{2}u^{-1/2}\]

Step 3: Find the Derivative of u with respect to x

Determine the derivative of \( u \) with respect to \( x \), denoted \( \frac{du}{dx} \). Since \( u = \sin x \), we use the derivative formula for sine:\[\frac{du}{dx} = \cos x\]

Step 4: Apply the Chain Rule

Use the chain rule to find \( \frac{dy}{dx} \). Multiply \( \frac{dy}{du} \) by \( \frac{du}{dx} \):\[\frac{dy}{dx} = \frac{1}{2} u^{-1/2} \cdot \cos x\]

Step 5: Substitute u back in terms of x

Substitute \( u = \sin x \) back into the derivative expression:\[\frac{dy}{dx} = \frac{1}{2} (\sin x)^{-1/2} \cdot \cos x\]This simplifies to:\[\frac{dy}{dx} = \frac{\cos x}{2 \sqrt{\sin x}}\]

Key Concepts

Function Derivatives

Understanding the concept of derivatives is crucial in calculus. A derivative is a measure of how a function changes as its input changes. In the context of the problem, we're looking at two functions: \( y = \sqrt{u} \) and \( u = \sin x \). We want to find out how the overall function \( y \) changes with respect to \( x \). This involves differentiating step by step: first \( y \) with respect to \( u \), and then \( u \) with respect to \( x \). These individual derivatives are then combined to find the overall rate of change, which is \( \frac{dy}{dx} \). This approach is essential in calculus whenever a function is composed of other functions.

• \( y \) as a function of \( u \)
• \( u \) as a function of \( x \)
• Overall derivative \( \frac{dy}{dx} \)

Power Rule

The power rule is a basic differentiation rule used widely in calculus when dealing with functions of the form \( x^n \). It states that if \( y = x^n \), then the derivative \( \frac{dy}{dx} = nx^{n-1} \). In our example, we have \( y = \sqrt{u} = u^{1/2} \). By applying the power rule, we determine \( \frac{dy}{du} \).
Simply put:

• Original function: \( u^{1/2} \)
• Applying power rule: \( \frac{d}{du}(u^{1/2}) = \frac{1}{2} u^{-1/2} \)

Using the power rule efficiently simplifies finding derivatives of similar functions. It gives a straightforward route to understanding how polynomial-like functions behave when they undergo changes.

Sine Derivative

The derivative of the sine function is fundamental in calculus. For \( u = \sin x \), the derivative \( \frac{du}{dx} = \cos x \) follows from the standard rules of differentiation for trigonometric functions. The sine function's derivative tells us the rate at which the sine of an angle changes with respect to that angle (in radians). The

• Sine function: \( \sin x \)
• Derivative using: \(\frac{d}{dx}(\sin x) = \cos x \)

This relationship between the sine and its derivative is an essential tool when analyzing the motion and oscillations you're likely to encounter in fields such as physics or engineering.

Chain Rule Application

The chain rule is an essential technique in calculus, allowing the differentiation of composite functions. When we have a situation where one function is nested inside another, the chain rule provides a systematic way to find the derivative. In this example, we have \( y = \sqrt{u} \) and \( u = \sin x \). To find \( \frac{dy}{dx} \), we employ the chain rule: multiply the derivative of \( y \) with respect to \( u \) by the derivative of \( u \) with respect to \( x \).

• Derivative of outer function: \( \frac{dy}{du} = \frac{1}{2}u^{-1/2} \)
• Derivative of inner function: \( \frac{du}{dx} = \cos x \)
• Application of chain rule: \( \frac{dy}{dx} = \frac{1}{2} u^{-1/2} \cdot \cos x \)

Finally, substituting \( u = \sin x \) back, we obtain \( \frac{dy}{dx} = \frac{\cos x}{2 \sqrt{\sin x}} \). This derivative reflects the chain rule's power in linking the behavior of composite functions.