Find dy/dx- y=arccos[(2^x+1)/(1+4^x)]

We'll have to differentiate the given function with
respect to x. Since the expression of the function is the result of composition of more
functions, we'll apply the chain rule.

dy/dx =
d{arccos[(2^x+1)/(1+4^x)]}/dx

(arccos u)' =
-u'/sqrt(1-u^2)

Let u = [(2^x+1)/(1+4^x)] => u' =
[(2^x+1)'*(1+4^x)-(2^x+1)*(1+4^x)]']/(1+4^x)^2

u' = [2^x*ln
2*(1 + 4^x) - 4^x*ln 4*(2^x + 1)]/(1+4^x)^2

u' = (2^x*ln2 +
2^3x*ln2-2^(3x+1)*ln2-2^(x+1)*ln2)/(1+4^x)^2

u' =
2^x*ln2(1+2^2x) - 2^(x+1)*ln2*(1+2^2x)

u' =
-2^x*ln2*(1+2^2x)

dy/dx =
[2^x*ln2*(1+2^2x)]/sqrt{1-[(2^x+1)/(1+4^x)]^2}

dy/dx =
[2^x*ln2*(1+2^2x)*(1+4^x)]]/sqrt
[2^x*(2^x-1)*(2+2^x+2^2x)]

The requested
derivative of the given function y is: dy/dx = [2^x*ln2*(1+2^2x)*(1+4^x)]]/sqrt
[2^x*(2^x-1)*(2+2^x+2^2x)].