Y = square of root a - square of X, what is the derivative of this function Please write down the specific steps,

Y = square of root a - square of X, what is the derivative of this function Please write down the specific steps,

y=√(a^2-x^2)
y'=1/[2√(a^2-x^2)]*(-2x)=-x/√(a^2-x^2)

Find the derivative of function y = (1 + cos2x) 2

y′=2（1+cos2x）（1+cos2x）′
=2（1+cos2x）（-sin2x）（2x）′
=4（1+cos2x）（-sin2x）
=-4sin2x-2sin4x

Let y = 1 / 2 cos2x, what is the derivative of Y Product of 1 / 2 and cos2x

y'=(1/2)(-sin2x 2)=-sin2x

Y = ㏑ derivative of cos2x There is 1-2sin ^ 2x = cos2x in the formula According to this formula (cos2x) '= - 2Sin ^ 2x, so Why is (cos2x) '= - 2sin2x in the answer?

(cos2x)'=(-2sin^2x)'
=-2*(sin ² x)'
=-2*2sinx*(sinx)'
=-2*(2sinxcosx)
=-2sin2x

What is the third derivative of SiNx?

The derivative of SiNx to the third power?
Or SiNx for cubic derivative?
The former, 3cosx (SiNx) ^ 2
The latter, - cosx

Find the n-th derivative of SiNx, please give me a process

(sinx)'=cosx=sin(x+π/2)
(sinx)''=[sin(x+π/2)]'=cos[x+(π/2)]=sin[x+2(π/2)]
……
（sinx)^(n)=[sin(x+(n-1)(π/2))]'=cos[x+(n-1)(π/2)]=sin[x+n(π/2)]