How do you differentiate $f(x) =arccos(2x + 1)$ ?

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Answer 1 · 2017-06-13T10:45:26

$f'(x)=-1/sqrt(-x^2-x).$

Recall that, $d/dt arc cos t=-1/sqrt(1-t^2).$

Therefore, $f(x)-arc cos(2x+1)$

$rArr f'(x)=-1/sqrt{1-(2x+1)^2}*d/dx(2x+1),...[because," The Chain Rule]."$

$rArr f'(x)=-2/sqrt(-4x^2-4x)=-1/sqrt(-x^2-x).$

How do you differentiate f(x) =arccos(2x + 1) f(x) =arccos(2x + 1) ?