Given that a is a real number, f (x) = (x2-4) (x-a); (1) find the derivative f '(x); (2) if f' (- 1) = 0, find the maximum and minimum value of F (x) on [- 2,2]; (3) if f (x) is increasing on (- ∞, - 2) and (2, + ∞), find the value range of A

Given that a is a real number, f (x) = (x2-4) (x-a); (1) find the derivative f '(x); (2) if f' (- 1) = 0, find the maximum and minimum value of F (x) on [- 2,2]; (3) if f (x) is increasing on (- ∞, - 2) and (2, + ∞), find the value range of A

(1) We get f (x) = x3-ax2-4x + 4a, f '(x) = 3x2-2ax-4 from the original formula. (2) we get a = 12 from F' (- 1) = 0, where f (x) = (x2-4) (X-12), f '(x) = 3x2-x-4. We get x = 43 or x = - 1 from F' (x) = 0, and f (43) = - 5027, f (- 1) = 92, f (- 2) = 0, f (2) = 0, so f (x) is in

The square of X + the square of y = 3

V square = 3

Y = (x ^ 5-x ^ 3 + 1) ^ 2 for derivative

Solution
y‘=2(x^5-x^3+1)(x^5-x^3+1)'
=2(x^5-x^3+1)(5x^4-3x^2)

F (x) = sin2x + cosx, derivation

Derivation of (sin2x) / X

[（sin2x）/x ]'
=(2xcos2x-sin2x)/x^2
(using the formula (U / V) '= (u'v-uv') / V ^ 2)

Derivation of F (sin2x)

2cos2x f'（sin2x）

Derivative y = SiNx. Find y ''
Calculate y = SiNx. Find y ''

y=sinx
y'=cosx
y"=(cosx)'=-sinx

Derivative of y = 5 ^ SiNx

Finding the n-order derivative of y = INX

y=Inx
y'=x^(-1)
y''=-x^(-2)
y'''=2x^(-3)
y''''=-3!x^(-4)
N-order derivative of y = INX = (- 1) ^ (n-1) * (n-1)! X ^ (- n)

Second derivative y = cos ^ 2x. INX + e ^ (- x ^ 2)

y'=- 2cosxsinx*lnx+(1/x)cos²x- 2xe^(-x²)=-sin2x*lnx+(1/x)cos²x- 2xe^(-x²)y''=-2cos2x*lnx- (1/x)sin2x- (1/x²)cos²x- (1/x)sin2x- 2e^(-x²)+ 4x²e^(-x²)=- 2cos2...