Question

F(x) = cos⁴3x, хо= π/9

Answer 1

Объяснение:

как смогла решить(если не правильно сори)

Answer 2

To find the derivative of the function $F(x) = \cos^4(3x)$ at $x_0 = \frac{\pi}{9}$, you can use the chain rule and the power rule for differentiation. The chain rule states that if you have a composition of functions, $f(g(x))$, then the derivative is $f'(g(x)) \cdot g'(x)$.

In this case, let $u = 3x$, so $F(x) = \cos^4(u)$. Now, you can apply the chain rule:

$$F'(x) = \frac{d}{dx}\left(\cos^4(u)\right) \cdot \frac{du}{dx}$$

Now, find the derivatives:

1. $\frac{d}{dx}\left(\cos^4(u)\right)$: To find the derivative of $\cos^4(u)$, you can use the power rule. The derivative of $\cos^n(u)$ with respect to $u$ is $-n\cos^{n-1}(u) \sin(u)$. In this case, $n = 4$, so:

$$\frac{d}{dx}\left(\cos^4(u)\right) = -4\cos^3(u) \sin(u)$$

2. $\frac{du}{dx}$: The derivative of $3x$ with respect to $x$ is simply 3.

Now, plug these values back into the chain rule formula:

$$F'(x) = \left(-4\cos^3(u) \sin(u)\right) \cdot 3$$

Now, we need to evaluate this derivative at $x_0 = \frac{\pi}{9}$, so plug in $u = 3x_0$:

$$F'\left(\frac{\pi}{9}\right) = \left(-4\cos^3\left(\frac{\pi}{3}\right) \sin\left(\frac{\pi}{3}\right)\right) \cdot 3$$

Now, evaluate the trigonometric functions at $\frac{\pi}{3}$:

$$\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$$

$$\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$$

Plug these values in:

$$F'\left(\frac{\pi}{9}\right) = \left(-4\left(\frac{1}{2}\right)^3 \cdot \frac{\sqrt{3}}{2}\right) \cdot 3$$

Now, simplify the expression:

$$F'\left(\frac{\pi}{9}\right) = -\left(\frac{1}{4}\right) \cdot \left(\frac{\sqrt{3}}{2}\right) \cdot 3$$

Multiply the numbers:

$$F'\left(\frac{\pi}{9}\right) = -\frac{3\sqrt{3}}{8}$$