Question

Find the derivative of the following functions. $$y=x \cos x \sin x$$

Short answer

Question: Find the derivative of the function $$y = x \cos x \sin x$$. Answer: The derivative of the function is $$y'(x)=\cos x \sin x - x \sin^2 x + x \cos^2 x$$.

Step-by-step solution

Identify the functions and their derivatives

We need to find the derivatives of the three functions: 1. $$u(x) = x$$, so its derivative is $$u'(x) = 1$$ 2. $$v(x) = \cos x$$, so its derivative is $$v'(x) = -\sin x$$ 3. $$w(x) = \sin x$$, so its derivative is $$w'(x) = \cos x$$

Apply the product rule for three functions

The product rule for three functions is as follows: $$(uvw)' = u'vw + uv'w + uvw'$$ In our case, the function $$y(x) = uvw$$, and we will substitute the function and its derivatives into the equation. $$y'(x) = (1)(\cos x)(\sin x) + (x)(-\sin x)(\sin x) + (x)(\cos x)(\cos x)$$

Simplify the expression

Simplify the expression to get the final derivative. $$y'(x)=\cos x \sin x - x \sin^2 x + x \cos^2 x$$ So, the derivative of the function $$y = x \cos x \sin x$$ is: $$y'(x)=\cos x \sin x - x \sin^2 x + x \cos^2 x$$

Key concepts

Product Rule

The product rule is an essential tool in calculus used for finding the derivative of the product of two or more functions. It's especially useful when you can't simplify a function into its individual components for easier differentiation. For two functions, the product rule states that if you have functions \( u(x) \) and \( v(x) \), then the derivative of their product is given by:

• \((uv)' = u'v + uv'\)

This means that the derivative of the product is the derivative of the first function times the second function, plus the first function times the derivative of the second. When dealing with three functions, the formula extends to:

• \((uvw)' = u'vw + uv'w + uvw'\)

Here, you differentiate each function one by one, keeping the others constant, and then add the results. In our given problem, applying the product rule helps streamline the differentiation process for the function \( y = x \cos x \sin x \). By treating each term according to the rule, you can tackle even the most complicated of expressions.

Trigonometric Functions

Trigonometric functions such as \( \cos x \) and \( \sin x \) are fundamental in mathematics, especially in calculus. These functions relate the angles of a triangle to the lengths of its sides and have various properties used in differentiation and integration. When differentiating trigonometric functions:

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

These results are crucial when applying differentiation techniques to mathematical expressions involving trigonometric functions. Recognizing the derivatives of these functions allows you to quickly apply rules like the product rule and simplify complex expressions. By observing these foundational derivatives, differentiating combinations of trigonometric functions becomes more manageable.

Differentiation Techniques

Differentiation techniques allow you to find the rate at which something changes, and they play a pivotal role in calculus. Beyond the basic rules of differentiation such as the power, constant, and sum rules, more specialized techniques include the product rule, quotient rule, and chain rule. This diversity lets you tackle a wide range of functions and combinations effectively. In this exercise, we used:

• The product rule to differentiate a product of three functions.
• Knowledge of basic trigonometric derivatives.

Other common techniques include:

• The chain rule, used when functions are composed of other functions.
• The quotient rule, applicable for ratios of functions.

By mastering these techniques, you can approach calculus problems methodically, breaking down each component function, and applying the appropriate rule. This structured approach ensures you handle each piece accurately, leading to a correct overall solution. These methods are crucial tools in your mathematical toolkit, allowing you to solve intricate problems efficiently.