Derivative Of F(x) = (6sin(x) + 5cos(x))arctan(x)
Hey guys! In this article, we're diving into a super interesting calculus problem: finding the derivative of the function f(x) = (6sin(x) + 5cos(x))arctan(x). This might look a bit intimidating at first, but don't worry, we'll break it down step by step. We're going to use some fundamental calculus rules, especially the product rule, and also remember the derivatives of trigonometric functions and the inverse tangent function. So, grab your notebooks, and let's get started!
Understanding the Function
Before we jump into the differentiation process, let's take a moment to really understand the function we're dealing with. Our function is f(x) = (6sin(x) + 5cos(x))arctan(x). Notice that it's a product of two main parts: the first part is a trigonometric expression, (6sin(x) + 5cos(x)), and the second part is the inverse tangent function, arctan(x), also commonly written as tan⁻¹(x). Recognizing this structure is crucial because it tells us we'll definitely need to use the product rule when we find the derivative.
The trigonometric part of our function combines sine and cosine functions, each multiplied by a constant. Remember, sine and cosine are the bread and butter of trigonometry, oscillating smoothly between -1 and 1. The constants 6 and 5 simply scale these oscillations. On the other hand, the arctangent function, arctan(x), gives us the angle whose tangent is x. It's the inverse operation of the tangent function, and its derivative has a neat and tidy form, which we'll use later. Visualizing these components separately can help us anticipate how they'll interact when we differentiate the whole function.
Why This Matters
Understanding the function's components helps us strategize our approach. We know we'll need the derivatives of sin(x), cos(x), and arctan(x). We also know the product rule is our main tool here. Breaking down the problem into smaller, manageable parts makes it less daunting and more fun, trust me!
Essential Calculus Rules and Derivatives
Okay, before we actually compute the derivative of our function, let's quickly review the essential calculus rules and derivatives that we'll be using. This is like making sure we have all the right tools in our toolbox before starting a DIY project. The most important rule for this problem is the product rule. Remember, the product rule helps us find the derivative of a function that is formed by the product of two other functions. If we have a function like f(x) = u(x)v(x), where u(x) and v(x) are both functions of x, then the product rule states that:
f'(x) = u'(x)v(x) + u(x)v'(x)
In simpler terms, 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 function. Got it? Great!
Now, let's talk about the specific derivatives we'll need. We have trigonometric functions and the inverse tangent function in our original function, so we need to know their derivatives. Here they are:
- The derivative of sin(x) is cos(x). (d/dx [sin(x)] = cos(x))
- The derivative of cos(x) is -sin(x). (d/dx [cos(x)] = -sin(x))
- The derivative of arctan(x) (or tan⁻¹(x)) is 1 / (1 + x²). (d/dx [arctan(x)] = 1 / (1 + x²))
These are fundamental derivatives that are super useful to memorize. They pop up all the time in calculus problems. With these rules and derivatives in our arsenal, we're well-equipped to tackle our main problem.
Why These Rules Are Key
The product rule is essential because our function is literally a product of two functions. And knowing the derivatives of trigonometric functions and the inverse tangent function is non-negotiable for this problem. Without these, we'd be stuck! So, make sure you're comfortable with these before moving on.
Applying the Product Rule
Alright, guys, now for the fun part – actually applying the product rule to find the derivative of our function f(x) = (6sin(x) + 5cos(x))arctan(x). Remember how we identified that this function is a product of two parts? That's our cue to bring in the product rule! Let's break it down step by step to keep things clear and organized.
First, we need to identify our u(x) and v(x). Let's set:
- u(x) = 6sin(x) + 5cos(x)
- v(x) = arctan(x)
Now, we need to find the derivatives of u(x) and v(x). This is where knowing those basic derivatives comes in handy. Let's start with u'(x):
u'(x) = d/dx [6sin(x) + 5cos(x)]
We can differentiate term by term. The derivative of 6sin(x) is 6cos(x) (since the derivative of sin(x) is cos(x)), and the derivative of 5cos(x) is -5sin(x) (since the derivative of cos(x) is -sin(x)). So:
u'(x) = 6cos(x) - 5sin(x)
Next up, we need v'(x), which is the derivative of arctan(x). Lucky for us, we know this one:
v'(x) = d/dx [arctan(x)] = 1 / (1 + x²)
Now that we have u(x), v(x), u'(x), and v'(x), we can plug them into the product rule formula:
f'(x) = u'(x)v(x) + u(x)v'(x)
Substituting our values, we get:
f'(x) = (6cos(x) - 5sin(x))arctan(x) + (6sin(x) + 5cos(x))(1 / (1 + x²))
And there you have it! We've applied the product rule and found the derivative. Now, let's simplify it a bit to make it look nicer.
Why This Method Works
The product rule allows us to handle functions that are products of other functions, which is exactly what we had here. By breaking the function into two parts and finding their individual derivatives, we could systematically apply the rule and find the overall derivative. It's like a puzzle – each piece fits together to give us the solution.
Simplifying the Derivative
Okay, so we've got our derivative: f'(x) = (6cos(x) - 5sin(x))arctan(x) + (6sin(x) + 5cos(x))(1 / (1 + x²)). It looks a bit clunky, right? Let's clean it up and make it more presentable. Simplifying derivatives not only makes them look better but can also make them easier to work with in future calculations. Plus, a simplified answer just feels more satisfying, doesn't it?
Looking at our expression, we have two main terms. The first term, (6cos(x) - 5sin(x))arctan(x), is already fairly simple. There's not much we can do to it without expanding it, and in this case, expansion doesn't really help. So, let's leave that one as is.
The second term, (6sin(x) + 5cos(x))(1 / (1 + x²)), can be rewritten to look a bit cleaner. We can think of the (1 / (1 + x²)) part as a fraction and multiply it with the expression in the parentheses. This gives us:
(6sin(x) + 5cos(x)) / (1 + x²)
Now, let's put it all together. Our simplified derivative is:
f'(x) = (6cos(x) - 5sin(x))arctan(x) + (6sin(x) + 5cos(x)) / (1 + x²)
This is a much cleaner and more manageable form of the derivative. We haven't changed the mathematical meaning, just the appearance. It's like tidying up your room – same room, but way more pleasant to be in!
Why Simplify?
Simplifying a derivative isn't just about aesthetics. A simplified expression is easier to use if you need to find critical points, analyze the function's behavior, or perform further calculus operations. It also reduces the chance of making errors in subsequent steps. So, always take a moment to simplify your results – your future self will thank you!
Final Answer and Conclusion
Alright, guys, we've reached the end of our journey! Let's recap what we've done. We started with the function f(x) = (6sin(x) + 5cos(x))arctan(x) and were tasked with finding its derivative. This looked like a challenge, but we broke it down step by step, used the product rule, and simplified our result.
After applying the product rule, we got:
f'(x) = (6cos(x) - 5sin(x))arctan(x) + (6sin(x) + 5cos(x))(1 / (1 + x²))
Then, we simplified it to get our final answer:
f'(x) = (6cos(x) - 5sin(x))arctan(x) + (6sin(x) + 5cos(x)) / (1 + x²)
This is the derivative of our original function. We tackled a problem that involved trigonometric functions, an inverse trigonometric function, and the product rule – that's quite a feat! Remember, the key to success in calculus (and in many other areas of life) is to break complex problems into smaller, manageable steps.
Key Takeaways
- Product Rule: Master the product rule – it's essential for differentiating functions that are products of other functions.
- Derivatives of Trig Functions: Know your derivatives of trigonometric functions (sin(x), cos(x), etc.) and inverse trig functions (arctan(x), etc.).
- Simplify: Always simplify your results. It makes them easier to work with and less prone to errors.
So, there you have it! We've successfully found the derivative of a complex function. I hope you found this explanation helpful and maybe even a little fun. Keep practicing, keep breaking problems down, and you'll become a calculus whiz in no time! If you have any questions or want to dive deeper into other calculus topics, let me know. Happy calculating!