Finding F'(π) For F(x) = 2x²cos(x - Π/2)

Hey guys! Today, we're diving into a fun calculus problem where we need to find the derivative of a function and then evaluate it at a specific point. Our function is $f(x) = 2x^2 \cos(x - \frac{\pi}{2})$, and we want to find the value of $f'(\pi)$. This means we first need to find the derivative of $f(x)$, which is $f'(x)$, and then plug in $\pi$ for $x$. Let's break it down step by step!

Understanding the Function and the Goal

Before we jump into the calculations, let's take a moment to understand what we're working with. Our function, $f(x) = 2x^2 \cos(x - \frac{\pi}{2})$, is a product of two simpler functions: $2x^2$ and $\cos(x - \frac{\pi}{2})$. This is a crucial observation because it tells us we'll need to use the product rule when we find the derivative. Remember, the product rule helps us differentiate functions that are multiplied together.

Our goal is clear: find $f'(\pi)$. This involves two main steps:

1. Find the derivative of $f(x)$, which is $f'(x)$.
2. Evaluate $f'(x)$ at $x = \pi$. This means substituting $\pi$ for $x$ in the expression for $f'(x)$ and simplifying.

With a clear understanding of the function and our goal, we're ready to start the differentiation process.

Step 1: Finding the Derivative f'(x)

To find the derivative $f'(x)$, we'll use the product rule. The product rule states that if we have a function $h(x) = u(x)v(x)$, then its derivative is given by:

$h'(x) = u'(x)v(x) + u(x)v'(x)$

In our case, we can identify:

• $u(x) = 2x^2$
• $v(x) = \cos(x - \frac{\pi}{2})$

Now, we need to find the derivatives of $u(x)$ and $v(x)$ separately.

Finding u'(x)

The derivative of $u(x) = 2x^2$ is straightforward. We use the power rule, which states that if $f(x) = ax^n$, then $f'(x) = nax^{n-1}$. Applying this to $u(x)$, we get:

$u'(x) = 2 * 2x^{2-1} = 4x$

So, the derivative of $u(x)$ is $4x$. This is a simple yet important step in applying the product rule correctly.

Finding v'(x)

Finding the derivative of $v(x) = \cos(x - \frac{\pi}{2})$ requires a bit more attention. We'll use the chain rule here. The chain rule is used when we have a composite function, meaning a function within a function. In our case, we have the cosine function with an inner function of $x - \frac{\pi}{2}$.

The chain rule states that if $g(x) = f(h(x))$, then $g'(x) = f'(h(x)) * h'(x)$.

Applying the chain rule to $v(x)$, we get:

• The derivative of the outer function, $\cos(x)$, is $-\sin(x)$.
• The derivative of the inner function, $x - \frac{\pi}{2}$, is $1$.

Therefore,

$v'(x) = -\sin(x - \frac{\pi}{2}) * 1 = -\sin(x - \frac{\pi}{2})$

So, the derivative of $v(x)$ is $- \sin(x - \frac{\pi}{2})$. This is another key component for using the product rule.

Applying the Product Rule

Now that we have $u(x)$, $v(x)$, $u'(x)$, and $v'(x)$, we can finally apply the product rule:

$f'(x) = u'(x)v(x) + u(x)v'(x)$

Substituting our expressions, we get:

$f'(x) = (4x) \cos(x - \frac{\pi}{2}) + (2x^2) (- \sin(x - \frac{\pi}{2}))$

Simplifying, we have:

$f'(x) = 4x \cos(x - \frac{\pi}{2}) - 2x^2 \sin(x - \frac{\pi}{2})$

So, we've found the derivative of our function: $f'(x) = 4x \cos(x - \frac{\pi}{2}) - 2x^2 \sin(x - \frac{\pi}{2})$. This is a big step, and now we're ready for the final part of the problem.

Step 2: Evaluating f'(π)

Now that we have the expression for $f'(x)$, we need to evaluate it at $x = \pi$. This means we'll substitute $\pi$ for $x$ in the expression we just found:

$f'(\pi) = 4(\pi) \cos(\pi - \frac{\pi}{2}) - 2(\pi)^2 \sin(\pi - \frac{\pi}{2})$

Let's simplify this step by step.

Simplifying the Trigonometric Functions

First, let's simplify the arguments of the cosine and sine functions:

$\pi - \frac{\pi}{2} = \frac{\pi}{2}$

So, we have:

$f'(\pi) = 4\pi \cos(\frac{\pi}{2}) - 2\pi^2 \sin(\frac{\pi}{2})$

Now, we need to evaluate the cosine and sine functions at $\frac{\pi}{2}$.

• $\cos(\frac{\pi}{2}) = 0$
• $\sin(\frac{\pi}{2}) = 1$

Substituting these values, we get:

$f'(\pi) = 4\pi (0) - 2\pi^2 (1)$

Final Calculation

Now we just need to do the arithmetic:

$f'(\pi) = 0 - 2\pi^2$

Therefore,

$f'(\pi) = -2\pi^2$

So, the value of $f'(\pi)$ is $-2\pi^2$. We've successfully found the derivative and evaluated it at the given point!

Conclusion

Awesome job, guys! We've walked through the process of finding the derivative of a function using the product and chain rules, and then we evaluated it at a specific point. Remember, the key to these problems is to break them down into smaller, manageable steps.

Here’s a quick recap of what we did:

1. Identified the function as a product and decided to use the product rule.
2. Found the derivatives of the individual components using the power rule and chain rule.
3. Applied the product rule to find $f'(x)$.
4. Substituted $\pi$ for $x$ in $f'(x)$.
5. Simplified the expression to find the final answer, $f'(\pi) = -2\pi^2$.

Keep practicing, and you'll become a calculus pro in no time! If you have any questions, feel free to ask. Keep up the great work!