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2 câu trả lời 214
Để tính đạo hàm của hàm số y=cot(4−x2), ta thực hiện như sau:
Sử dụng quy tắc đạo hàm của hàm hợp: (f(g(x)))′=f′(g(x))⋅g′(x)
Ở đây, ta có:
* f(u)=cot(u)
* g(x)=4−x2
Ta biết rằng:
* dducot(u)=−csc2(u)
* ddx(4−x2)=−2x
Áp dụng quy tắc hàm hợp:
dydx=ddxcot(4−x2)=−csc2(4−x2)⋅(−2x)=2xcsc2(4−x2)
Vậy, đạo hàm của y=cot(4−x2) là:
y′=2xcsc2(4−x2)
Sure! Let's analyze the two functions you've provided: y=tan(2x) and y=cot(4−x2).
c) y=tan(2x)
1. Domain:
tan(2x) is undefined where cos(2x)=0⇒2x=π2+kπ(k∈Z)⇒x=π4+kπ2(k∈Z).
Therefore, the domain of y=tan(2x) is:
x∈R∖{π4+kπ2:k∈Z}.
2. Range:
The range is all real numbers:
y∈R.
3. Period:
The function tan(kx) has a period of πk. Therefore, for y=tan(2x):
Period=π2.
4. Graph:
The graph of tan(2x) will have vertical asymptotes at x=π4+kπ2. The function will oscillate between −∞ and +∞ between each pair of asymptotes.
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d) y=cot(4−x2)
1. Domain:
The cotangent function is undefined wherever sin(4−x2)=0, which happens when:
4−x2=nπ(n∈Z)⇒x2=4−nπ⇒x=±√4−nπ.
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∈R.
3. Behavior:
Since y=cot(4−x2) has vertical asymptotes at points where 4−x2=nπ, the function will have a pattern of oscillation between these asymptotes.
4. Graph:
The graph of y=cot(4−x2) 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 π2 and has vertical asymptotes at specific points based on kπ/2.
- For y=cot(4−x2), it has vertical asymptotes determined by solving 4−x2=nπ 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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