Find derivative y = (x-1) ² （x+1） ³

Find derivative y = (x-1) ² （x+1） ³

Y = f (x) * g (x), then take the derivative to get y '= f' (x) * g (x) + F (x) * g '(x). Obviously, y = (x-1) ² (x+1) ³ Then we get the derivative y '= [(x-1) of Y ²]' * (x+1) ³ + (x-1) ² * [(x+1) ³]'= 2(x-1) *(x+1) ³ + (x-1) ² * ...

Y = 1 / X (2 + 5x) ^ 10 derivation

Let a = x (5x + 2) ^ 10, B = 5x + 2, then the original formula = 1 / A, a = x * B ^ 10, DB = 5DX
Derivative (1 / a) '= D (1 / a) / DX = (- 1 / A ^ 2) (DA / DX)
=(-1/A^2)[d(x*B^10)/dx]
=(-1/A^2)[(B^10dx+xdB^10)/dx]
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Function y = cos (2x + π) 2) An axis of symmetry equation of the image is () A. x=-π two B. x=-π four C. x=π eight D. x=π

The symmetry axis equation of this function is 2x + π
two  ＝ K π (K ∈ z), when k = 0, x =   −π
4．
Therefore, B

Given the function cos (2a pi / 3) + sin (a-pi / 4) sin (a + pi / 4), find the minimum positive period of the function and the equation of image symmetry axis Find the minimum positive period of the function and the equation of the image symmetry axis Find the value range of the function on the interval [- pi / 12, PI / 2]

Y=cos(2a-π/3）+sin(a-π/4)sin(a+π/4)= cos(2a-π/3）+ sin(a-π/4)sin[(a-π/4)+π/2]= cos(2a-π/3）+ sin(a-π/4)cos(a-π/4)= cos(2a-π/3）+1/2*sin(2a-π/2)= cos(2a-π/3）-1/2*cos(2a)= cos(2a) cosπ/3+...

Known function y = 3sin (2x + π) 6) , then its axis of symmetry equation is () A. x=0 B. x=-π twelve C. x=π six D. x=π three

By 2x + π
6=kπ+π
2, get x = k π
2+π
6（k∈Z），
Let k = 0 and get x = π
6，
Its axis of symmetry equation is x = π
6，
Therefore: C

What is the symmetry axis equation of the image with function y = cos (2x + 1 / 2)? What is the center of symmetry

The axis of symmetry equation is cos (2x + 1 / 2) = ± 1
2x+1/2=k∏
x=k∏/2-1/4
The center of symmetry is cos (2x + 1 / 2) = 0
2x+1/2= ∏/2+k∏
x=-1/4+∏/4+k∏/2