How does 1-cos (x + y) in y = sin (x + y) derivation come from

How does 1-cos (x + y) in y = sin (x + y) derivation come from

Derivative of X
Then y '= cos (x + y) * (x + y)'
y'=cos(x+y)*(1+y')
So y '= cos (x + y) + y'cos (x + y)
Y '[1-cos (x + y)] = cos (x + y)
y'=cos(x+y)/[1-cos(x+y)]

x=a( θ－ sin θ) y=a(1-cos θ) Take the derivative of the parametric equation, and the answer is sin θ/ 1-cos θ

Derivative dy / DX
Parametric equation first pair θ Derivation
dx=a(1-cos θ) d θ , dy=a(0+sin θ) d θ= asin θ d θ
Compared with the above two formulas
dy/dx= asin θ d θ / a(1-cos θ) d θ = sin θ/ (1-cos θ)

Y = cos ^ 2 (x ^ 2-x). The answer is - 2cos (x ^ 2-x) sin (x ^ 2-x) (2x-1)

y=cos ² (x ²- x)
y'=2cos(x ²- x) × [cos(x ²- x)]' × (x ²- x)'
y'=2cos(x ²- x)[-sin(x ²- x)](2x-1)
y'=2(1-2x)sin(x ²- x)cos(x ²- x)

Is sin π x + cos π x a periodic function and how to find its period

Is a periodic function. By summing the difference product, it is equivalent to csin (pi * x + pi / 4), where C is a constant. Then its period is 2pi / pi = 2

Given the function f (x) = x ^ 3 + (1-A) x ^ 2-A (a + 2) x + B, if the function f (x) is not monotonic in the interval (- 1,1), find the value range of A. answer (- 5,1 / 2) ∪ (1 / 2,1) My method is to find out the derivative of F (x), and then just make f '(1) * f' (- 1) < 0, but the calculation is (- 5, - 1). Is there any mistake in this idea? If so, where is the mistake?

⒈ 1 and - 1 are not in the function definition domain, so the derivatives of these two points can not be used in principle. 2. The function is not monotonous, and the derivative values at both ends may not be different signs, for example, sin (x) is on [0,2 π],

Known function f (x) = 1 3x3 − (4m − 1) x2 + (15m2 − 2m − 7) x + 2 is an increasing function on (- ∞, + ∞), so the value range of M is () A. M ＜ - 4 or m ＞ - 2 B. -4＜m＜-2 C. 2＜m＜4 D. M ＜ 2 or m ＞ 4

For f (x) = 1
The derivative of 3x3 − (4m − 1) x2 + (15m2 − 2m − 7) x + 2 is obtained
f′（x）=x2-2（4m-1）x+（15m2-2m-7）
Known function f (x) = 1
3x3 − (4m − 1) x2 + (15m2 − 2m − 7) x + 2 is an increasing function on (- ∞, + ∞)
Therefore, f ′ (x) ＞ 0
That is, find the value range of m that makes x2-2 (4m-1) x + (15m2-2m-7) > 0
It can be seen that the function opening is upward so that △ < 0
Solve [- 2 (4m-1)] 2-4 (15m2-2m-7) ＜ 0, and obtain
2＜m＜4
So choose C