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

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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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