Question

Đã trả lời bởi chuyên gia

Answer 1

Để tính đạo hàm của hàm số \(y = \cot(4 - x^2)\), ta thực hiện như sau:

Sử dụng quy tắc đạo hàm của hàm hợp: \((f(g(x)))' = f'(g(x)) \cdot g'(x)\)

Ở đây, ta có:

* \(f(u) = \cot(u)\)
* \(g(x) = 4 - x^2\)

Ta biết rằng:

* \(\frac{d}{du} \cot(u) = -\csc^2(u)\)
* \(\frac{d}{dx} (4 - x^2) = -2x\)

Áp dụng quy tắc hàm hợp:

\[
\frac{dy}{dx} = \frac{d}{dx} \cot(4 - x^2) = -\csc^2(4 - x^2) \cdot (-2x) = 2x \csc^2(4 - x^2)
\]

Vậy, đạo hàm của \(y = \cot(4 - x^2)\) là:

\[
y' = 2x \csc^2(4 - x^2)
\]

Answer 2

Sure! Let's analyze the two functions you've provided: \(y = \tan(2x)\) and \(y = \cot(4 - x^2)\).

c) \(y = \tan(2x)\)

1. Domain:
\[
\tan(2x) \text{ is undefined where } \cos(2x) = 0 \, \Rightarrow \, 2x = \frac{\pi}{2} + k\pi \, (k \in \mathbb{Z}) \, \Rightarrow \, x = \frac{\pi}{4} + \frac{k\pi}{2} \, (k \in \mathbb{Z}).
\]
Therefore, the domain of \(y = \tan(2x)\) is:
\[
x \in \mathbb{R} \setminus \left\{ \frac{\pi}{4} + \frac{k\pi}{2} : k \in \mathbb{Z} \right\}.
\]

2. Range:
The range is all real numbers:
\[
y \in \mathbb{R}.
\]

3. Period:
The function \(\tan(kx)\) has a period of \(\frac{\pi}{k}\). Therefore, for \(y = \tan(2x)\):
\[
\text{Period} = \frac{\pi}{2}.
\]

4. Graph:
The graph of \(\tan(2x)\) will have vertical asymptotes at \(x = \frac{\pi}{4} + \frac{k\pi}{2}\). The function will oscillate between \(-\infty\) and \(+\infty\) between each pair of asymptotes.

---

d) \(y = \cot(4 - x^2)\)

1. Domain:
The cotangent function is undefined wherever \(\sin(4 - x^2) = 0\), which happens when:
\[
4 - x^2 = n\pi \, (n \in \mathbb{Z}) \Rightarrow x^2 = 4 - n\pi \Rightarrow x = \pm\sqrt{4 - n\pi}.
\]
Therefore, the domain consists of all real numbers except those points, and it depends on the values of \(n\).

2. Range:
The range is all real numbers, similar to the tangent function.
\[
y \in \mathbb{R}.
\]

3. Behavior:
Since \(y = \cot(4 - x^2)\) has vertical asymptotes at points where \(4 - x^2 = n\pi\), the function will have a pattern of oscillation between these asymptotes.

4. Graph:
The graph of \(y = \cot(4 - x^2)\) will have a series of oscillations, with each segment being similar to the cotangent function.

Summary:

- For \(y = \tan(2x)\), the function is periodic with a period of \(\frac{\pi}{2}\) and has vertical asymptotes at specific points based on \(k\pi/2\).
- For \(y = \cot(4 - x^2)\), it has vertical asymptotes determined by solving \(4 - x^2 = n\pi\) and oscillates between those asymptotes.

If you need specific calculations, like derivatives or values at certain points, please let me know!

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Answer 3