What is the LN (- x) derivative? What is the derivative of F (x) = ax ln (- x)?

What is the LN (- x) derivative? What is the derivative of F (x) = ax ln (- x)?

[ln(-x)]'
=[1/(-x)]*(-x)'
=-[1/(-x)]
=1/x
So f '(x) = A-1 / X

Let f '(x) be continuous on [a, b], prove: Lim（ λ →+∞)∫(a,b)f(x)cos( λ x)dx=0

Using partial integral
∫(a,b)f(x)cos( λ x)dx=1/ λ * ∫(a,b)f(x)dsin( λ x)
=1/ λ * { [f(x)sin( λ x)] |(a,b) - ∫(a,b)f'(x)sin( λ x)dx}
Because f '(x) is continuous on [a, b]
0≤|∫(a,b)f'(x)sin( λ x) Dx|≤∫ (a, b) |f '(x) |dx = a (and λ (independent constant)
Similarly, we can analyze [f (x) sin（ λ x) ] | (a, b) is also a bounded quantity
So Lim（ λ →+∞) ∫(a,b)f(x)cos( λ x)dx=lim( λ →+∞)1/ λ * { [f(x)sin( λ x)] |(a,b) - ∫(a,b)f'(x)sin( λ x)dx]}=0
The reason is that infinitesimal quantity times bounded quantity

Let f (x) have a continuous first derivative at x = 1, f '(1) = 2, find d [f (COS √ (x-1))] / DX when Lim X - > 1 +, why is the answer not 2

d[f(cos√(x-1))]/dx=f'(x)*(-sin√(x-1))*1/2*1/√(x-1)=-1/2*f'(x)*sin√(x-1)/√(x-1)
D [f (COS √ (x-1))] / DX = Lim X - > 1 + [- 1 / 2 * f '(x) * sin √ (x-1) / √ (x-1)] = - 1 / 2 * 2 * 1 = - 1 when Lim X - > 1 +
The key is the limit of sin √ (x-1) / √ (x-1) = 1

Find the derivative of COS ^ 3 (3x)?

{cos^3 (3x)}'
= 3 cos^2(3X) {cos(3x)}'
= 3 cos^2(3X) {-sin(3x)} (3x)'
= 3 cos^2(3X) {-sin(3x)} 3
= -9 sin(3x) cos^2(3x)

Find the derivative of 4sin (3x) cos (x) Such as title

12cos(3x)cos(x)-4sin(3x)sin(x)

How to find the derivative of COS (3x)?

(cos(3x))'=-sin(3x)*(3x)'=-3sin(3x)