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

giorgiana1976 | Student

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)].

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