Problem 11 - Calculus

Question

Find the derivatives of the functions in \(1-26\). $$ x^{1 / 2} \sin ^{2} x+(\sin x)^{1 / 2} $$

Step-by-Step Solution

Answer

The derivative is \(\frac{1}{2}x^{-1/2} \sin^2 x + 2x^{1/2} \sin x \cos x + \frac{\cos x}{2\sqrt{\sin x}}\).

Step 1: Identify the components of the function

The function is composed of two parts: \ 1. \(x^{1/2} \sin^2 x\) \ 2. \((\sin x)^{1/2}\) \ We'll find the derivative of each part separately and then add them together.

Step 2: Differentiate the first component

The first component is \(x^{1/2} \sin^2 x\). Use the product rule: \\(\frac{d}{dx}(u \cdot v) = u'v + uv'\), where \(u = x^{1/2}\) and \(v = \sin^2 x\). \ - \(u' = \frac{1}{2}x^{-1/2}\) \ - \(v = (\sin x)^2\) and \(v' = 2\sin x \cos x\) \ The derivative is: \ \[\frac{d}{dx}(x^{1/2} \sin^2 x) = \frac{1}{2}x^{-1/2} \cdot \sin^2 x + x^{1/2} \cdot 2\sin x \cos x\] \ Simplify it to: \ \[\frac{1}{2}x^{-1/2} \sin^2 x + 2 \cdot x^{1/2} \cdot \sin x \cos x\]

Step 3: Differentiate the second component

The second component is \((\sin x)^{1/2}\). Use the chain rule: \ \(\frac{d}{dx}(u^n) = nu^{n-1} \cdot u'\), where \(u = \sin x\) and \(n = 1/2\). \ - \(u' = \cos x\) \ The derivative is: \ \[\frac{1}{2}(\sin x)^{-1/2} \cdot \cos x \]This simplifies to: \ \[\frac{\cos x}{2\sqrt{\sin x}}\]

Step 4: Combine the derivatives

Add the derivatives from Steps 2 and 3 to find the derivative of the complete function: \\[\frac{1}{2}x^{-1/2} \sin^2 x + 2x^{1/2} \sin x \cos x + \frac{\cos x}{2\sqrt{\sin x}}\] This is the derivative of the original function.

Key Concepts

Product Rule

In calculus, the product rule is used to find the derivative of a function that is the product of two other functions. When you have a function in the form of \( f(x) = u(x) \cdot v(x) \), the product rule states that its derivative \( f'(x) \) is given by:

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

This rule is essential when working with functions like \( x^{1/2} \sin^2 x \). Here, \( u(x) = x^{1/2} \) and \( v(x) = \sin^2 x \).
To apply the product rule:

• First, find \( u'(x) \), the derivative of \( u(x) \). For \( u(x) = x^{1/2} \), we get \( u'(x) = \frac{1}{2}x^{-1/2} \).
• Next, find \( v'(x) \), the derivative of \( v(x) = \sin^2 x \). Using the chain rule (which we'll cover next), \( v'(x) = 2 \sin x \cos x \).
• Finally, substitute \( u'(x) \) and \( v'(x) \) back into the product rule formula to get the derivative of the product.

By using the product rule, you effectively break down the differentiation process into manageable parts.

Chain Rule

The chain rule is a fundamental tool in calculus used for differentiating compositions of functions. It states that if you have a function inside another function, the derivative of the compositions \( f(g(x)) \) is:

• \( (f(g(x)))' = f'(g(x)) \cdot g'(x) \)

This becomes especially helpful when differentiating functions like \((\sin x)^{1/2}\), where you have a power of a trigonometric function.
To apply the chain rule:

• Identify the outer function \( f(u) = u^{1/2} \) and the inner function \( u(x) = \sin x \).
• Differentiate the outer function: \( f'(u) = \frac{1}{2}u^{-1/2} \).
• Differentiate the inner function: \( u'(x) = \cos x \).
• Apply the chain rule: \((\sin x)^{1/2}' = \frac{1}{2}(\sin x)^{-1/2} \cdot \cos x \).

Using the chain rule simplifies finding derivatives of nested functions, turning a complex problem into basic steps.

Trigonometric Functions

Trigonometric functions, such as sine and cosine, are vital in many calculus problems. These functions have their own set of rules for differentiation, which are used extensively when working with derivatives that involve them.

• The derivative of \( \sin x \) is \( \cos x \).
• The derivative of \( \cos x \) is \( -\sin x \).

These rules make it straightforward to tackle the differentiation of functions involving these trigonometric elements, like \( \sin^2 x \) and \( \cos x \).
For example, the derivative of \( \sin^2 x \), using the product and chain rules, becomes \( 2 \sin x \cos x \).

• This result is from recognizing \( \sin^2 x \) as a product of \( \sin x \cdot \sin x \) and then applying the product and chain rules to simplify the derivative.

Understanding how to differentiate trigonometric functions is crucial, as these functions frequently appear in various calculus problems.