How do you find the derivative of $y=arcsin(2x+1)$ ?

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Answer 1 · 2016-09-19T19:53:41

$dy/dx=1/sqrt(-x^2-x)$

Note that:

$d/dxarcsin(x)=1/sqrt(1-x^2)$

So, according to the chain rule:

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

So, where $f(x)=2x+1$ , and $f'(x)=2$ :

$dy/dx=d/dxarcsin(2x+1)=2/sqrt(1-(2x+1)^2)$

$=2/sqrt(1-(4x^2+4x+1))$

$=2/sqrt(-4x^2-4x)$

$=2/(sqrt4sqrt(-x^2-x))$

$=1/sqrt(-x^2-x)$

How do you find the derivative of y=arcsin(2x+1)y=arcsin(2x+1)?