Problem 26 - Calculus

Question

Find the derivatives of the functions in \(1-26\). $$ x \sin x+\cos x $$

Step-by-Step Solution

Answer

The derivative is \(x \cos x\).

Step 1: Identify the Functions

The function given is composed of two functions: \[ f(x) = x \sin x + \cos x \]This is a sum of the product of two functions \(x\) and \(\sin x\), and another function \(\cos x\). We'll need to differentiate each part separately.

Step 2: Apply the Product Rule

The first part of the function is \( x \sin x \). To differentiate this, apply the product rule:\[ \text{If } u(x) = x \text{ and } v(x) = \sin x, \text{ then } u'(x) = 1 \text{ and } v'(x) = \cos x. \]The product rule is given by: \[ (uv)' = u'v + uv'. \]Therefore, the derivative of \(x \sin x\) is:\[ (x \sin x)' = 1 \cdot \sin x + x \cdot \cos x = \sin x + x \cos x. \]

Step 3: Differentiate \(\cos x\)

The derivative of \(\cos x\) is a standard derivative:\[ (\cos x)' = -\sin x. \]

Step 4: Combine the Results

Now, combine the two derivatives derived in Steps 2 and 3:\[ f'(x) = (x \sin x)' + (\cos x)' = (\sin x + x \cos x) + (-\sin x). \]

Step 5: Simplify the Expression

Combine like terms in the derivative:\[ f'(x) = \sin x + x \cos x - \sin x. \]The \(\sin x\) and \(-\sin x\) terms cancel each other out:\[ f'(x) = x \cos x. \]

Key Concepts

Product Rule Explanation

In the world of calculus derivatives, the product rule is one of the fundamental techniques used when dealing with the differentiation of products of two functions. If you have two functions, say \( u(x) \) and \( v(x) \), the product rule states that the derivative of their product is not simply the product of their derivatives. Instead, it is expressed as: \[ (uv)' = u'v + uv'. \] This means you take the derivative of the first function and multiply it by the second function, and then add to it the first function times the derivative of the second function.

• Differentiate \( u(x) \) to find \( u'(x) \).
• Replace \( u'(x) \) while keeping \( v(x) \) constant.
• Then differentiate \( v(x) \) to obtain \( v'(x) \).
• Finally, replace \( v'(x) \) while keeping \( u(x) \) constant.

It can be remembered with the mnemonic: "one primes one, times the other and plus one, times the other primed". It's very helpful for working with products of simple to more complex functions.

Understanding Trigonometric Functions

Trigonometric functions, like \( \sin x \) and \( \cos x \), are fundamental in calculus because they often appear in functions that need to be differentiated. Learning their derivatives from the beginning is crucial for solving various calculus problems.

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

These derivatives are derived from the unit circle definition of these functions. It's important to memorize these, as they frequently arise. When we encounter a function involving returns to trigonometric components, like in this problem, treating them individually and using rules like product or chain rule is the standard approach. In calculus, remember that the trigonometric identities can also sometimes be useful in simplifying complex expressions after differentiation.

Simplifying Derivatives

Once you've applied the necessary rules and calculated derivatives, like with the product rule or for trigonometric functions, the next step is simplification. This is crucial in making the derivative easy to understand and further use. In our problem, after applying the product and finding the derivative of the trigonometric function, we combined the results and noticed that some terms could cancel out: \[ f'(x) = \sin x + x \cos x - \sin x. \] Here, \( \sin x \) and \(-\sin x \) cancel each other, simplifying to: \[ f'(x) = x \cos x. \]

• Combining like terms helps reduce complexity.
• Removing unnecessary components leaves a cleaner result.

Simplification is often missed, but it's an essential part of obtaining the final tidy form of a derivative, providing clarity in both homework and further applications.