Find H'(-5) Given F(x) And H(x) Derivatives

Hey guys, let's dive into a cool math problem where we're dealing with derivatives. We've got a function ${ f(x) }$ that's differentiable, which basically means we can find its derivative at any point. Then, we have another function ${ h(x) }$ defined in terms of ${ f(x) }$. Our mission, should we choose to accept it, is to find the value of the derivative of ${ h(x) }$ at ${ x = -5 }$, denoted as ${ h'(-5) }$. Sounds like fun, right? Let's break it down step by step.

Understanding the Problem

Before we jump into calculations, let's make sure we understand what we're given. We know that:

• ${ f(x) }$ is a differentiable function. This is crucial because it allows us to use the rules of calculus.
• ${ h(x) = [f(x)]^2 }$. This tells us that ${ h(x) }$ is the square of the function ${ f(x) }$.
• ${ f(-5) = 1 }$. This means that when we plug ${ x = -5 }$ into ${ f(x) }$, we get 1.
• ${ f'(-5) = -1 }$. This is the derivative of ${ f(x) }$ evaluated at ${ x = -5 }$, and it equals -1. Remember, the derivative gives us the instantaneous rate of change of the function.

Our goal is to find ${ h'(-5) }$, which is the derivative of ${ h(x) }$ evaluated at ${ x = -5 }$. To do this, we'll need to use the chain rule, a fundamental concept in calculus.

Applying the Chain Rule

The chain rule is our best friend when we're dealing with composite functions—functions within functions. In our case, ${ h(x) }$ is a composite function because it's ${ f(x) }$ squared. The chain rule states that if we have a function ${ h(x) = g(f(x)) }$, then its derivative ${ h'(x) }$ is given by:

${ h'(x) = g'(f(x)) imes f'(x) }$

In simpler terms, we take the derivative of the outer function ${ g }$ evaluated at the inner function ${ f(x) }$, and then multiply it by the derivative of the inner function ${ f'(x) }$. This might sound a bit complicated, but it's easier than it looks once we apply it to our specific problem.

For our function ${ h(x) = [f(x)]^2 }$, we can think of the outer function as ${ g(u) = u^2 }$ and the inner function as ${ f(x) }$. So, let's find the derivatives:

• The derivative of ${ g(u) = u^2 }$ is ${ g'(u) = 2u }$.
• The derivative of ${ f(x) }$ is ${ f'(x) }$, which we already know is ${ -1 }$ at ${ x = -5 }$.

Now, we can apply the chain rule:

${ h'(x) = g'(f(x)) imes f'(x) = 2f(x) imes f'(x) }$

This formula tells us how to find the derivative of ${ h(x) }$ at any point ${ x }$. But we're specifically interested in ${ x = -5 }$, so let's plug that in.

Calculating h'(-5)

Now that we have the formula for ${ h'(x) }$, we can find ${ h'(-5) }$ by plugging in ${ x = -5 }$:

${ h'(-5) = 2f(-5) imes f'(-5) }$

We know that ${ f(-5) = 1 }$ and ${ f'(-5) = -1 }$, so we can substitute these values:

${ h'(-5) = 2(1) imes (-1) }$

${ h'(-5) = -2 }$

And there we have it! The derivative of ${ h(x) }$ evaluated at ${ x = -5 }$ is ${ -2 }$.

Putting It All Together

Let's recap what we've done. We started with a function ${ h(x) }$ defined as the square of another function ${ f(x) }$, and we were given some information about ${ f(x) }$ and its derivative at ${ x = -5 }$. Our goal was to find ${ h'(-5) }$. To do this, we used the chain rule, which is essential for differentiating composite functions. We found the derivative of ${ h(x) }$ in terms of ${ f(x) }$ and ${ f'(x) }$, and then we plugged in the given values to find ${ h'(-5) = -2 }$.

Importance of Differentiability

It's important to highlight why the fact that ${ f(x) }$ is differentiable is so crucial. Differentiability means that the function has a derivative at every point in its domain. In simpler terms, the graph of the function has a well-defined tangent line at each point. This allows us to use the rules of calculus, like the chain rule, to find derivatives. If ${ f(x) }$ weren't differentiable, we wouldn't be able to find ${ f'(-5) }$, and the whole problem would fall apart.

Understanding the Chain Rule Further

The chain rule can be a bit tricky to grasp at first, but it's one of the most powerful tools in calculus. Think of it as a way to peel an onion—you differentiate the outer layer first, then move to the inner layers. In our case, the outer layer was the squaring function, and the inner layer was ${ f(x) }$. By applying the chain rule, we were able to systematically find the derivative of the composite function.

Real-World Applications

Derivatives might seem like abstract mathematical concepts, but they have tons of real-world applications. They're used in physics to describe velocity and acceleration, in economics to model marginal cost and revenue, and in computer science for machine learning algorithms. Understanding derivatives is crucial for anyone working in these fields.

Practice Makes Perfect

If you're still a bit unsure about derivatives and the chain rule, don't worry! The best way to master these concepts is through practice. Try working through more examples, and don't be afraid to ask for help if you get stuck. Math can be challenging, but it's also incredibly rewarding when you finally understand something new.

Conclusion

So, there you have it! We've successfully found ${ h'(-5) }$ using the chain rule. This problem highlights the importance of understanding fundamental calculus concepts and how they can be applied to solve interesting problems. Keep practicing, and you'll become a derivative pro in no time! Remember, math is like a muscle—the more you use it, the stronger it gets. And who knows? Maybe you'll be the one solving the next big math problem that changes the world!