For x > 1, if (2x)²ʸ = 4e²ˣ⁻ʸ, then (1 + loge2x)² dy/dx is equal to:

(2x)²ʸ = 4e²ˣ⁻ʸ
2y ln(2x) = ln4 + 2x - y
2y ln2x + y = ln4 + 2x
y(2 ln2x + 1) = 2x + ln4
y = (2x + ln4) / (1 + 2 ln2x)
dy/dx = [2(1 + 2 ln2x) - (2x + ln4)(2/x)] / (1 + 2 ln2x)²
dy/dx = [2 + 4ln2x - 4 - (2ln4)/x] / (1 + 2ln2x)²
dy/dx = [4ln2x - 2 - (2ln4)/x] / (1 + 2ln2x)²
(1 + loge2x)² dy/dx = (1 + 2ln2x)² dy/dx = 4ln2x - 2 - (2ln4)/x
Let's simplify:
y = (2x + 2ln2)/(1 + 2ln2x)
dy/dx = [2(1+2ln2x) - (2x+2ln2)(4/x)]/(1+2ln2x)²
dy/dx = [2+4ln2x - 8 - 8ln2/x]/(1+2ln2x)²
(1+2ln2x)² dy/dx = 4ln2x - 6 - 8ln2/x
If (2x)²ʸ = 4e²ˣ⁻ʸ, then taking logarithms on both sides, we have:
2y ln(2x) = ln4 + 2x - y
2y ln(2x) + y = ln4 + 2x
y (2 ln(2x) + 1) = 2x + ln4
y = (2x + ln4) / (1 + 2 ln(2x))
Differentiating with respect to x:
dy/dx = [(1 + 2 ln(2x))(2) - (2x + ln4)(4/x)] / (1 + 2 ln(2x))²
dy/dx = [2 + 4 ln(2x) - 8 - (4 ln4)/x] / (1 + 2 ln(2x))²
dy/dx = [4 ln(2x) - 6 - (4 ln4)/x] / (1 + 2 ln(2x))²
(1 + ln(2x))² dy/dx = (1 + 2 ln(2x))² dy/dx = 4 ln(2x) - 6 - (4 ln 4)/x = 4 ln(2x) - 6 - (8 ln 2)/x
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