How to find the derivation of (1-2x) to the eighth power

How to find the derivation of (1-2x) to the eighth power

y=(1-2x)^8
Let t = 1-2x
y=t^8
y'=(t')*(t^8)'
=-2*8(1-2x)^7
=-16(1-2x)^7

Derivation of (2x + 1) to the 100th power

100*(2X+1)^99*2=200*(2X+1)^99
(derivative of composite function, 2x + 1 has to be solved again)

3、 Find the derivative or differential 1) y = (2x-1) cubic, y ′. 2) y = arctan2x, find y ′. 3) f (x) = 5x quartic-3x? + 2x + 1, find F

1 y'=6(2x-1)^2
2 y'=2/(1+(2x)^2)
3 y'=20x^3+6x+2

To the 2x power of 2, what is the derivative equal to

[2^(2*x)]'=[2^(2*x)]*ln2*(2*x)'=[2^(2*x+1)]*ln2

What is the derivative of (2x + 3) to the fifth power

(2x+3)^5=5*2(2x+3)^4=10(2x+3)^4

Y = u to the power of 2x, how much is it,

Is u a known quantity?
y=u^2x
y′=(u^2x)′=2x(u)^(2x-1)

How to do the 2x power derivative of X? What's the solution?

X ^ 2x = e ^ (2xlnx) so: (x ^ 2x) '= [e ^ (2xlnx)] = e ^ (2xlnx) * x ^ 2x = e ^ (2xlnx) therefore: (x ^ 2x)' = [e ^ (2xlnx)] = e ^ (2xlnx) * [2lnx + 2] = x ^ 2x * (2lnx + 2) is the key to obtain the function in the form of e ^ X

Y = e to the power of (sin ^ 2 times 2x) I don't know how to write clearly. I hope you can help me

y'=[sin^2(2x)]' * e^[sin^2(2x)]
y'=[sin(2x)]' * 2sin(2x) * e^[sin^2(2x)]
y'=(2x)'cos(2x) * 2sin(2x) * e^[sin^2(2x)]
y'=2 sin(4x) * e^[sin^2(2x)]

What is the 2007 power of 2 minus the 2009 power of 2 and the 2008 power of 2 / 2

2^2008/(2^2007-2^2009)
=2^2008/[2^2007(1-2^2)]
=(-1/3)*2^2008/2^2007
=(-1/3)*2
=-2/3

Calculate the 2008 power of (- 2) and the 2009 power of (- 2)

Original formula = 2 ^ 2008-2 ^ 2009
=-2^2008