How do you find the derivative of $y = arcsin(3x + 2)$ ?

Question

3087 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-18T12:08:59

The answer is $=3/sqrt(1-(3x+2)^2)$

If $y=arcsin(3x+2)$

Then,

$siny=3x+2$

$dy/dxcosy=3$

$dy/dx=3/cosy$

But,

$cos^2y+sin^2y=1$

$cos^2y=1-(3x+2)^2$

$cosy=sqrt(1-(3x+2)^2)$

Therefore,

$dy/dx=3/sqrt(1-(3x+2)^2)$

How do you find the derivative of y = arcsin(3x + 2)y = arcsin(3x + 2)?