Derivatives Of Inverse Trigonometric Functions Step-by-Step Solutions

Hey there, math enthusiasts! Today, we're diving deep into the fascinating world of inverse trigonometric functions and how to find their derivatives. If you've ever wondered how to differentiate functions like arctan(x) or arccos(x), you're in the right place. We'll break down the process step-by-step, making it super easy to grasp. Let's tackle two interesting examples that will help solidify your understanding. So, grab your pencils, and let's get started!

6. 1 Delving into the Derivative of f(x) = arctan(√x)

Let's kick things off with our first function: f(x) = arctan(√x). This might look a bit intimidating at first, but trust me, it's totally manageable. The key here is to remember the chain rule. This rule is your best friend when you're dealing with composite functions – functions within functions. In our case, we have the arctangent function (arctan) acting on the square root of x (√x). So, before we even start, let's remember the derivative of the arctangent function. The derivative of arctan(u) with respect to u is 1 / (1 + u^2). Keep this in your back pocket; we're going to need it.

Now, let's apply the chain rule. The chain rule essentially says that the derivative of a composite function is the derivative of the outer function (in our case, arctan) evaluated at the inner function (√x), multiplied by the derivative of the inner function. Sounds like a mouthful, right? Let's break it down. First, we find the derivative of the outer function, arctan(u), with respect to u, which we already know is 1 / (1 + u^2). But instead of 'u', we plug in our inner function, √x. This gives us 1 / (1 + (√x)^2), which simplifies to 1 / (1 + x). Next up, we need to find the derivative of the inner function, √x. Remember that √x can also be written as x^(1/2). Using the power rule, which states that the derivative of x^n is n*x^(n-1), the derivative of x^(1/2) is (1/2) * x^((1/2) - 1), which simplifies to (1/2) * x^(-1/2). This can be further rewritten as 1 / (2√x). Finally, we multiply the derivative of the outer function evaluated at the inner function by the derivative of the inner function. So, we have (1 / (1 + x)) * (1 / (2√x)). And there you have it! We've found the derivative. To clean it up a bit, we can write the final answer as 1 / (2√x(1 + x)). See? Not so scary after all!

Wrapping Up the Arctangent Derivative: Let's recap quickly what we've done. We identified the composite function, remembered the derivative of the arctangent function, applied the chain rule by finding the derivatives of the outer and inner functions, and then multiplied them together. This systematic approach is key to tackling any composite function. The arctan(√x) example is a classic one that highlights the power and elegance of the chain rule. By understanding this method, you're well-equipped to handle a wide range of derivative problems involving inverse trigonometric functions. Remember, practice makes perfect, so try working through similar examples to really nail down the concept. You've got this!

6. 2 Unraveling the Derivative of h(t) = √t * arccos(e^t)

Now, let's move on to our second example: h(t) = √t * arccos(e^t). This one is a bit more complex, but don't worry, we'll break it down piece by piece. Here, we have a product of two functions: √t and arccos(e^t). This means we'll need to use the product rule. The product rule states that the derivative of the product of two functions, say u(t) and v(t), is u'(t)v(t) + u(t)v'(t). In simpler terms, it's the derivative of the first function times the second function, plus the first function times the derivative of the second function. So, let's identify our u(t) and v(t). Here, u(t) = √t and v(t) = arccos(e^t). First things first, we need to find the derivative of each of these functions. We already know how to differentiate √t from our previous example. As we discussed, √t can be written as t^(1/2), and its derivative, using the power rule, is (1/2) * t^(-1/2), which simplifies to 1 / (2√t). So, u'(t) = 1 / (2√t).

Next, we need to find the derivative of v(t) = arccos(e^t). This is where things get interesting. We have another composite function! We'll need to use the chain rule again. First, let's recall the derivative of the arccosine function. The derivative of arccos(u) with respect to u is -1 / √(1 - u^2). Keep this formula handy. Now, applying the chain rule, we first find the derivative of the outer function, arccos(u), with respect to u, which is -1 / √(1 - u^2). We then plug in our inner function, e^t, for u, giving us -1 / √(1 - (e^t)2), which simplifies to -1 / √(1 - e^(2t)). Next, we need to find the derivative of the inner function, e^t. Luckily, the derivative of e^t with respect to t is simply e^t. Now, we multiply the derivative of the outer function evaluated at the inner function by the derivative of the inner function. So, we have (-1 / √(1 - e^(2t))) * e^t, which can be written as -e^t / √(1 - e^(2t)). That's the derivative of arccos(e^t)! Now that we have u(t), u'(t), v(t), and v'(t), we can finally apply the product rule. Remember, the product rule is u'(t)v(t) + u(t)v'(t). Plugging in our values, we get (1 / (2√t)) * arccos(e^t) + √t * (-e^t / √(1 - e^(2t))). This is our derivative! We can clean it up a bit to make it look nicer, but the key is that we've successfully applied both the product rule and the chain rule to find the derivative of this complex function.

Final Thoughts on the Arccosine Derivative: Whoa, that was quite a journey, wasn't it? Let's quickly recap what we did. We recognized the product of two functions, applied the product rule, used the chain rule to find the derivative of arccos(e^t), and then combined everything to get our final answer. This example really showcases how different differentiation rules can come together in a single problem. The √t * arccos(e^t) example is a fantastic exercise in combining the product rule and the chain rule. By mastering these techniques, you'll be able to tackle even the most challenging derivative problems. Keep practicing, keep exploring, and you'll become a derivative pro in no time!

Summing Up Our Trigonometric Derivative Adventure

So, guys, we've journeyed through the derivatives of two inverse trigonometric functions, and hopefully, you're feeling much more confident now. We tackled f(x) = arctan(√x) using the chain rule and then conquered h(t) = √t * arccos(e^t) by combining the product and chain rules. These examples are a testament to the power of understanding and applying the fundamental rules of calculus. Remember, the key to mastering derivatives is practice. Work through plenty of examples, and don't be afraid to make mistakes – that's how we learn! Inverse trigonometric functions might seem tricky at first, but with a solid understanding of the chain rule, product rule, and the derivatives of the basic inverse trig functions, you'll be differentiating like a pro in no time.

Keep exploring the world of calculus, and remember, every problem is just a puzzle waiting to be solved. You've got the tools, you've got the knowledge, now go out there and conquer those derivatives! And if you ever feel stuck, remember to revisit these examples and break down the problem step-by-step. Happy differentiating, everyone!