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

1 Answer

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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)].