Derivative Of F(x)/g(x) At X=1: A Calculus Problem

Hey Plastik Magazine readers! Ever get that feeling when a calculus problem just clicks? Today, we're diving into a cool derivative problem that combines the quotient rule with some given function values. It might seem daunting at first, but trust me, we'll break it down step-by-step so it’s super clear. So, buckle up and let's get our math on!

Problem Statement: Unpacking the Calculus Conundrum

Okay, so here’s the deal. We're given two functions: $f(x)$ and $g(x)$. We know that $f(1) = 0$ and $f'(1) = -7$. This is crucial information, so keep it in mind! The other function, $g(x)$, is defined as the square root of $x$, or $\sqrt{x}$. Our mission, should we choose to accept it (and we totally do!), is to find the derivative of the quotient $\frac{f(x)}{g(x)}$, and then evaluate that derivative at the point where $x = 1$. This means we need to first find the general formula for the derivative and then plug in $x = 1$ to get our final answer. Sounds like fun, right? The key to solving this problem lies in understanding and applying the quotient rule correctly. We also need to remember how to differentiate the square root function. Let's get started by revisiting the quotient rule and the derivative of $g(x)$. We'll then combine everything to tackle the main problem. Remember, in calculus, breaking down complex problems into smaller, manageable steps is always the best strategy. So, let's take a deep breath and start with the basics!

The Quotient Rule: Our Trusty Tool for Dividing Functions

Alright, let's talk about the quotient rule. This is our main weapon in tackling derivatives of fractions where both the numerator and denominator are functions of $x$. If you've seen it before, great! If not, no worries, we'll go through it together. The quotient rule states that if we have a function $h(x)$ defined as the quotient of two other functions, say $u(x)$ and $v(x)$, so $h(x) = \frac{u(x)}{v(x)}$, then the derivative of $h(x)$, denoted as $h'(x)$, is given by the following formula:

$$h'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2}$$

Don't let all those symbols scare you! Let's break it down in plain English. The derivative of the quotient is equal to "the denominator times the derivative of the numerator, minus the numerator times the derivative of the denominator, all divided by the denominator squared." Phew! That's a mouthful, but it’s the essence of the quotient rule. Now, why is this rule so important? Well, many functions we encounter in real life and in more advanced math are expressed as quotients. Think about rates of change, ratios, or even trigonometric functions (like tangent, which is sine divided by cosine). The quotient rule allows us to handle these complex expressions with confidence. In our particular problem, $f(x)$ will play the role of $u(x)$ and $g(x)$ will be $v(x)$. So, we'll need to find their derivatives, $f'(x)$ and $g'(x)$, respectively. We already have some information about $f(x)$ and $f'(x)$ at $x=1$, which is a good start. But what about $g(x)$? That's our next stop!

Differentiating g(x): Square Roots Aren't Scary!

Now, let’s tackle the derivative of $g(x) = \sqrt{x}$. You might remember this one, but let's refresh our memory just in case. Remember that the square root of $x$ can also be written as $x$ raised to the power of one-half, or $x^{1/2}$. This is a crucial step because it allows us to use the power rule for differentiation, which is a super handy tool. The power rule states that if we have a function of the form $x^n$, where $n$ is any real number, then its derivative is given by $nx^{n-1}$. In other words, we multiply by the exponent and then reduce the exponent by one. Applying the power rule to $g(x) = x^{1/2}$, we get:

$$g'(x) = \frac{1}{2}x^{\frac{1}{2} - 1} = \frac{1}{2}x^{-\frac{1}{2}}$$

Now, let's clean this up a bit. Remember that a negative exponent means we take the reciprocal. So, $x^{-1/2}$ is the same as $\frac{1}{x^{1/2}}$, which is also $\frac{1}{\sqrt{x}}$. Putting it all together, we have:

$$g'(x) = \frac{1}{2\sqrt{x}}$$

So, the derivative of $g(x) = \sqrt{x}$ is $g'(x) = \frac{1}{2\sqrt{x}}$. We've conquered another piece of the puzzle! Now we know both $g(x)$ and its derivative $g'(x)$. This is excellent progress. We're almost ready to apply the quotient rule to the whole expression $\frac{f(x)}{g(x)}$. Before we do that, let’s take a quick recap of what we've learned so far. We know the quotient rule, we know the derivative of the square root function, and we have the given values for $f(1)$ and $f'(1)$. Let's put it all together!

Applying the Quotient Rule: Putting It All Together

Okay, guys, this is where the magic happens! We're going to apply the quotient rule to our function $\frac{f(x)}{g(x)}$. Remember the quotient rule formula? It's:

$$\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{g(x)f'(x) - f(x)g'(x)}{[g(x)]^2}$$

We know $g(x) = \sqrt{x}$, $g'(x) = \frac{1}{2\sqrt{x}}$, and we're given $f(1) = 0$ and $f'(1) = -7$. Let's plug these into the formula. First, let's substitute the functions and their derivatives into the quotient rule:

$$\frac{d}{dx}\left[\frac{f(x)}{\sqrt{x}}\right] = \frac{\sqrt{x}f'(x) - f(x)\left(\frac{1}{2\sqrt{x}}\right)}{(\sqrt{x})^2}$$

Now, let's simplify this expression a bit. The denominator $(\sqrt{x})^2$ just becomes $x$. So we have:

$$\frac{d}{dx}\left[\frac{f(x)}{\sqrt{x}}\right] = \frac{\sqrt{x}f'(x) - \frac{f(x)}{2\sqrt{x}}}{x}$$

This looks a bit cleaner, right? Now, remember our ultimate goal: we want to evaluate this derivative at $x = 1$. So, let's plug in $x = 1$ and use the given values $f(1) = 0$ and $f'(1) = -7$:

Evaluating at x=1: The Grand Finale

Alright, folks, we're in the home stretch! It's time to plug in $x = 1$ into our derivative expression and see what we get. We have:

$$\frac{d}{dx}\left[\frac{f(x)}{\sqrt{x}}\right]\bigg|_{x=1} = \frac{\sqrt{1}f'(1) - \frac{f(1)}{2\sqrt{1}}}{1}$$

Now, let's substitute the values we know: $f(1) = 0$ and $f'(1) = -7$. Also, $\sqrt{1}$ is just 1. So, we have:

$$\frac{(1)(-7) - \frac{0}{2(1)}}{1}$$

This simplifies to:

$$\frac{-7 - 0}{1} = -7$$

And there you have it! The derivative of $\frac{f(x)}{g(x)}$ evaluated at $x = 1$ is -7. We did it! We took a complex calculus problem and broke it down into manageable steps. We used the quotient rule, differentiated a square root function, and plugged in given values. It’s all about taking it one step at a time and remembering the fundamental rules. So, high-five yourselves, you've conquered this derivative challenge!

Key Takeaways: What We Learned Today

So, what did we learn today, guys? We tackled a pretty cool calculus problem that involved finding the derivative of a quotient of two functions. We used the quotient rule, which is super important for differentiating fractions where both the numerator and denominator are functions of $x$. We also brushed up on our power rule skills to differentiate the square root function. And, most importantly, we saw how to combine these techniques with given function values to find a specific answer. The key takeaways from this problem are:

1. The Quotient Rule: Remember the formula! It's $\frac{d}{dx}\left[\frac{u(x)}{v(x)}\right] = \frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2}$. Practice using it, and it'll become second nature.
2. Power Rule and Square Roots: Don't forget that $\sqrt{x}$ is the same as $x^{1/2}$, and the power rule is your friend for differentiating these kinds of expressions.
3. Breaking Down Problems: Complex calculus problems can seem overwhelming, but break them down into smaller steps. Identify the rules you need to use, find the necessary derivatives, and then put it all together.
4. Using Given Information: Pay close attention to any given values or conditions in the problem. They're there to help you! In our case, $f(1) = 0$ and $f'(1) = -7$ were crucial for finding the final answer.

Calculus can be challenging, but it's also incredibly rewarding. Keep practicing, keep asking questions, and you'll become a derivative-master in no time! Until next time, keep those calculators handy and your minds sharp!