Hey math enthusiasts! Today, we're diving into the world of calculus to differentiate the function sin³x cos³x with respect to x. This might seem a bit daunting at first, but trust me, we'll break it down into manageable steps. We'll use the chain rule, product rule, and some trigonometric identities to simplify things. Let's get started, shall we?

Understanding the Problem: Differentiating sin³x cos³x

Differentiating sin³x cos³x is essentially finding the rate of change of this function as x varies. The function sin³x cos³x involves trigonometric functions raised to powers, so we'll need to use some powerful tools from calculus. Our main goal is to find d/dx (sin³x cos³x). Before we jump in, let's briefly review the key rules and concepts that we'll be using.

First, the chain rule. The chain rule is used when you have a composite function. If y = f(g(x)), then dy/dx = f'(g(x)) * g'(x). In simpler terms, you differentiate the outer function with respect to the inner function and multiply it by the derivative of the inner function. Next up, the product rule. The product rule is used when you have the product of two functions. If y = u(x)v(x), then dy/dx = u'(x)v(x) + u(x)v'(x). Basically, you differentiate the first function, multiply by the second, then add the first function multiplied by the derivative of the second. Finally, we need to know the derivatives of sin(x) and cos(x). We know that d/dx (sin x) = cos x and d/dx (cos x) = -sin x. We are also going to need the power rule: d/dx (x^n) = nx^(n-1)*. That's all the essentials we will need to know to solve this problem. These are the building blocks that will help us solve the problem. Now, are you ready to solve the derivative?

So, why is differentiating sin³x cos³x important? Well, it appears in several areas of physics and engineering, especially in problems related to oscillations and wave phenomena. For example, in the study of damped harmonic motion or the analysis of electrical circuits with sinusoidal sources, this function, or similar ones, may represent the energy or power associated with the system. Knowing how to differentiate such functions gives you a strong foundation to solve complex problems.

Step-by-Step Differentiation of sin³x cos³x

Alright, let's get our hands dirty! We need to find d/dx (sin³x cos³x). Since we have a product of two functions, sin³x and cos³x, we will have to use the product rule. Let:

• u(x) = sin³x
• v(x) = cos³x

So, the product rule states that d/dx (uv) = u'v + uv'. We'll find u' and v' first.

Finding the derivative of u(x) = sin³x

To find u', which is the derivative of sin³x, we'll use the chain rule. Remember that sin³x can be written as (sin x)³. Let g(x) = sin x. Then, u(x) = g(x)³. The chain rule is d/dx (g(x)³) = 3(g(x))² * g'(x). We know that the derivative of sin x is cos x. Thus:

• u'(x) = 3(sin x)² * cos x = 3sin²x cos x

Finding the derivative of v(x) = cos³x

Similarly, to find v', the derivative of cos³x, we will again use the chain rule. We can rewrite cos³x as (cos x)³. Let h(x) = cos x. Then, v(x) = h(x)³. Using the chain rule, d/dx (h(x)³) = 3(h(x))² * h'(x). We know that the derivative of cos x is -sin x. Therefore:

• v'(x) = 3(cos x)² * (-sin x) = -3cos²x sin x

Applying the Product Rule

Now that we have u' = 3sin²x cos x and v' = -3cos²x sin x, we can plug these back into the product rule formula:

• d/dx (sin³x cos³x) = u'v + uv'
• d/dx (sin³x cos³x) = (3sin²x cos x)(cos³x) + (sin³x)(-3cos²x sin x)
• d/dx (sin³x cos³x) = 3sin²x cos⁴x - 3sin⁴x cos²x

Simplifying the Result

Okay, we've got the derivative, but we can simplify it even further. We can factor out 3sin²x cos²x from both terms:

• 3sin²x cos⁴x - 3sin⁴x cos²x = 3sin²x cos²x(cos²x - sin²x)

Now, here's where a trigonometric identity comes in handy. We know that cos²x - sin²x = cos(2x). So, we can substitute this into our result. Therefore, the simplified derivative becomes:

• d/dx (sin³x cos³x) = 3sin²x cos²x cos(2x)

Alternative Simplification using Trigonometric Identities

Another way to simplify the derivative is by using the double-angle identity for sine and cosine. We know that sin(2x) = 2sin x cos x. Let's manipulate our derivative to use this identity. First, we have:

• 3sin²x cos²x cos(2x)

We can rewrite this as:

• (3/4)(2sin x cos x)² cos(2x)

Then, we know that sin(2x) = 2sin x cos x. So we substitute this in the formula and we get:

• (3/4)sin²(2x) cos(2x)

Both 3sin²x cos²x cos(2x) and (3/4)sin²(2x) cos(2x) are correct, and they are equivalent. They both represent the derivative of the original function. The choice of which form to use often depends on the context of the problem and what you are trying to do with the result. Both are correct and mathematically valid representations of the derivative.

Conclusion: The Final Answer

So, guys, after all that hard work, the derivative of sin³x cos³x with respect to x is 3sin²x cos²x cos(2x) or (3/4)sin²(2x) cos(2x). We've used the product rule, chain rule, and some trigonometric identities to get to the solution. It's a great example of how different calculus rules and identities come together to solve a problem. Great job!

This process is crucial not just for academic purposes but also for real-world applications. Being able to differentiate trigonometric functions like this is fundamental in fields such as physics (analyzing wave functions), engineering (signal processing), and even economics (modeling cyclical patterns). Understanding how these mathematical tools work opens doors to a deeper comprehension of the world around us. Keep practicing, and you'll find these problems become easier with time. Remember to break down complex problems into smaller, manageable steps, and always double-check your work. You've got this!

I hope you enjoyed this guide. Keep practicing, and happy calculating! If you have any questions, feel free to ask!