First Derivative Of F(x) = √(sin² 3x): Solved!

Hey everyone! Let's dive into a cool calculus problem today: finding the first derivative of the function $f(x) = \sqrt{sin^2 3x}$. This might look a bit intimidating at first, but don't worry, we'll break it down step by step so it's super easy to follow. We'll use the chain rule, which is a fundamental concept in calculus, to solve this. So, grab your pencils and notebooks, and let's get started!

Understanding the Function

Before we jump into the differentiation, let's make sure we really understand what the function $f(x) = \sqrt{sin^2 3x}$ is all about. Guys, this function is actually a composition of several simpler functions. Think of it like layers of an onion – we need to peel each layer back one at a time to differentiate it correctly.

• First, we have the innermost layer, which is $3x$. This is a linear function, nice and straightforward.
• Next, we take the sine of this, giving us $sin(3x)$. So far, so good!
• Then, we square the result, ending up with $sin^2(3x)$. We're getting closer!
• Finally, we take the square root of the whole thing, which brings us to $\sqrt{sin^2 3x}$. Phew, that’s the complete function!

Understanding this composition is crucial because it tells us we’ll need to use the chain rule. The chain rule is our best friend when dealing with composite functions, and it basically tells us how to differentiate them layer by layer. Remember, the chain rule states that if we have a function $f(g(x))$, its derivative is $f'(g(x)) \\cdot g'(x)$. We'll be applying this rule multiple times in our problem, so let’s keep it in mind.

Applying the Chain Rule

Okay, now for the fun part – actually finding the derivative! We'll tackle this step by step, using the chain rule to peel away those layers we identified earlier. Remember, the function is $f(x) = \sqrt{sin^2 3x}$.

Step 1: Rewrite the Square Root

First, let's rewrite the square root as a power of $\frac{1}{2}$. This makes it easier to apply the power rule later on. So, we rewrite $f(x)$ as:

$f(x) = (sin^2 3x)^{\frac{1}{2}}$

This small change makes a big difference in how we approach the derivative. It’s a neat little trick that simplifies things quite a bit. Trust me, you’ll be using this trick all the time in calculus!

Step 2: Differentiate the Outer Layer

Now, we start with the outermost layer, which is the power of $\frac{1}{2}$. Using the power rule, we bring down the exponent and subtract 1 from it. The power rule states that if $f(x) = x^n$, then $f'(x) = nx^{n-1}$. Applying this to our function, we get:

$\frac{1}{2}(sin^2 3x)^{-\frac{1}{2}}$

Notice that we've only differentiated the outermost layer so far. We still need to deal with the inner layers. This is where the chain rule really shines. We're not done yet, guys; we need to multiply by the derivative of the inside function.

Step 3: Differentiate the Inside Function

The inside function is $sin^2 3x$. This is itself a composite function, so we need to apply the chain rule again! This time, we'll focus on the square, which is the outer layer of this inner function. We rewrite $sin^2 3x$ as $(sin 3x)^2$.

Applying the power rule again, we bring down the 2 and subtract 1 from the exponent:

$2(sin 3x)^1 = 2 sin 3x$

But wait, there's more! We're still not at the innermost layer. We need to multiply by the derivative of the function inside the parentheses, which is $sin 3x$.

Step 4: Differentiate sin 3x

The derivative of $sin 3x$ is $cos 3x$, but we need to remember the chain rule again! We have one more layer to peel back: the $3x$ inside the sine function. The derivative of $3x$ with respect to $x$ is simply 3. So, the derivative of $sin 3x$ is:

$3 cos 3x$

We're finally at the end of the chain! We've differentiated each layer, one by one. Now, we just need to put it all together.

Combining the Derivatives

Now that we've found the derivatives of each layer, we need to multiply them together, as the chain rule instructs us. Let's recap what we found:

• Derivative of the outermost layer: $\frac{1}{2}(sin^2 3x)^{-\frac{1}{2}}$
• Derivative of $sin^2 3x$: $2 sin 3x$
• Derivative of $sin 3x$: $3 cos 3x$

Multiplying these together, we get:

$f'(x) = \frac{1}{2}(sin^2 3x)^{-\frac{1}{2}} \cdot 2 sin 3x \cdot 3 cos 3x$

Simplifying the Expression

Our expression looks a bit messy right now, so let's simplify it. We can cancel out the 2 in the numerator and denominator, and we can rewrite the negative exponent as a reciprocal square root:

$f'(x) = \frac{3 sin 3x cos 3x}{\sqrt{sin^2 3x}}$

This is already looking much cleaner! But we can simplify it even further. Notice that we have a $\sqrt{sin^2 3x}$ in the denominator. Assuming $sin 3x$ is positive (which is a common assumption in these types of problems), we can simplify this to $|sin 3x|$, or just $sin 3x$:

$f'(x) = \frac{3 sin 3x cos 3x}{sin 3x}$

Now, we can cancel out the $sin 3x$ terms:

$f'(x) = 3 cos 3x$

The Final Answer

So, after all that work, we've arrived at the final answer! The first derivative of $f(x) = \sqrt{sin^2 3x}$ is:

$f'(x) = 3 cos 3x$

Isn't that satisfying? We took a seemingly complex function and, by systematically applying the chain rule and simplifying, we found its derivative. Great job, guys!

Checking the Options

Now, let's take a look at the options provided and see which one matches our answer. The options were:

• $\frac{2}{3}cos^{-\frac{1}{3}}3xsin3x$
• $2 cot 3x \cdot \sqrt{sin^2 3x}$
• $2cos^{-\frac{1}{3}}3x$
• $\frac{2}{3}cos^{-\frac{1}{3}}3x$
• $-2 cot 3x \cdot \sqrt{sin^2 3x}$

None of these options directly match our answer of $3 cos 3x$. However, this doesn't necessarily mean we made a mistake. Sometimes, the options are given in a different form, and we need to manipulate our answer to see if it matches any of them.

Let's think about this. We simplified $\sqrt{sin^2 3x}$ to $sin 3x$, assuming $sin 3x$ is positive. If $sin 3x$ is negative, then $\sqrt{sin^2 3x}$ would be $-sin 3x$, and our derivative would be $-3 cos 3x$. This doesn't match any of the options either.

It seems there might be an issue with the provided options, or perhaps some further simplification is required that isn't immediately obvious. In a real test scenario, this would be a good point to double-check our work and, if confident in our solution, look for the closest match or consider if there's a typo in the options.

Key Takeaways

Before we wrap up, let's highlight some key takeaways from this problem:

• Chain Rule: The chain rule is essential for differentiating composite functions. Remember to work from the outermost layer inward, multiplying the derivatives of each layer.
• Simplification: Always simplify your expression as much as possible. This not only makes the answer cleaner but also helps in matching it with the given options.
• Rewriting Functions: Rewriting functions (like changing a square root to a power of $\frac{1}{2}$) can make differentiation much easier.
• Attention to Detail: Be meticulous with each step. Calculus problems often involve multiple steps, and a small error early on can lead to a completely wrong answer.

Practice Makes Perfect

Calculus can be tricky, but with practice, you'll become more comfortable and confident. Try working through similar problems, and don't be afraid to make mistakes – that's how we learn! The more you practice, the better you'll get at recognizing patterns and applying the right techniques.

So, there you have it! We've successfully found the first derivative of $f(x) = \sqrt{sin^2 3x}$. Keep practicing, and you'll be a calculus pro in no time. Thanks for joining me, and I'll see you in the next problem!