Problem 15 - Calculus

Question

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

Step-by-Step Solution

Answer

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

Step 1: Differentiate the First Term

The first term is \( \frac{1}{2}x^2 \sin x \). To differentiate it, we apply the product rule since it is a product of \( \frac{1}{2}x^2 \) and \( \sin x \). The product rule states that if \( u(x) \) and \( v(x) \) are functions, then \((uv)' = u'v + uv'\).\\Set \( u = \frac{1}{2}x^2 \) and \( v = \sin x \):\- \( u' = x \) since \( \frac{1}{2}x^2 \) differentiates to \( x \).\- \( v' = \cos x \) since \( \sin x \) differentiates to \( \cos x \).\\Putting it together: \[ \left( \frac{1}{2}x^2 \sin x \right)' = x \sin x + \frac{1}{2}x^2 \cos x \].

Step 2: Differentiate the Second Term

The second term of the expression is \( -x \cos x \). We use the product rule again. Here, \( u = x \) and \( v = \cos x \).\- \( u' = 1 \) since \( x \) differentiates to \( 1 \).\- \( v' = -\sin x \) since \( \cos x \) differentiates to \( -\sin x \).\\Applying the product rule: \[ (u v)' = u'v + uv' = (1)(\cos x) + (x)(-\sin x) = \cos x - x \sin x \]. Therefore, \( -(x \cos x)' = -\cos x + x \sin x \).

Step 3: Differentiate the Third Term

The third term is \( \sin x \). The derivative of \( \sin x \) is \( \cos x \).

Step 4: Combine the Derivatives

Now, we combine the derivatives of each term obtained from the previous steps:\- From Step 1: \( x \sin x + \frac{1}{2}x^2 \cos x \)\- From Step 2: \( -\cos x + x \sin x \)\- From Step 3: \( \cos x \)\\Adding these up, we get: \[ \left( \frac{1}{2}x^2 \sin x - x \cos x + \sin x \right)' = x \sin x + \frac{1}{2}x^2 \cos x - \cos x + x \sin x + \cos x \].

Step 5: Simplification

Combine like terms in the expression obtained in Step 4. Notice that \( \cos x \) and \(- \cos x \) cancel each other out, and \( x \sin x + x \sin x \) simplifies to \( 2x \sin x \).\\The simplified result is: \[ 2x \sin x + \frac{1}{2}x^2 \cos x \].

Key Concepts

Product Rule

In derivative calculus, the product rule is immensely useful, especially when you're dealing with the differentiation of products of two functions. It allows us to differentiate more complex functions by simplifying them into their component parts. The product rule is generally expressed as \[(uv)' = u'v + uv' \]where \( u \) and \( v \) are functions of \( x \). This means the derivative of \( u \) times \( v \) is the derivative of \( u \) multiplied by \( v \), plus \( u \) multiplied by the derivative of \( v \).

• Essentially, you take the derivative of the first function, multiply by the second, then take the derivative of the second and multiply by the first.
• This method must be applied individually to parts of an expression when they are multiplied together.

In the given example, the expression \( \frac{1}{2}x^2 \sin x \) is a product where \( u = \frac{1}{2}x^2 \) and \( v = \sin x \). The product rule helps us break down the differentiation into manageable steps.

Trigonometric Differentiation

Trigonometric differentiation involves finding the derivatives of trigonometric functions like \( \sin x \), \( \cos x \), and others. This is a frequent step in problems involving waves, oscillations, and circular motion. Here are the basic derivatives you need to remember:

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

When differentiating products that include these trigonometric functions, you often need to apply these fundamental derivatives in conjunction with the product rule.

In the exercise provided, each term containing \( \sin x \) or \( \cos x \) requires us to substitute these basic derivatives during the differentiation process. This ensures that you handle all parts of the equation correctly and efficiently.

Simplification of Expressions

Simplification comes in after all derivatives have been computed and collected. At times, the resulting expression from differentiation can look cluttered. Simplifying helps to organize and reduce the expression into a more compact form.

In simplification, you will:

• Combine like terms to reduce complexity.
• Cancel out terms that appear both positively and negatively in the expression.

For example, in the step-by-step solution we simplify \(-\cos x + x \sin x + \cos x\) to \(x \sin x\) by recognizing that \( \cos x \) and \(-\cos x\) cancel each other. Additionally, duplicating terms like \( x \sin x \) can be combined to read as \( 2x \sin x \).

Simplification not only eases the understanding of the expression but also ensures that it's in the simplest form possible for further analysis or application.