# Thread: Derivations of Inverse Trig Functions

1. ## Derivations of Inverse Trig Functions

My professor did not explain how to find the derivation of an inverse trig function very well, so I need some basic help in understanding what steps to take. I understand the derivations, and the need to use the chain rule and sometimes the product/quotient rule, but I don't understand these problems:

y = arcsin(2x +1)

Would this be the next step? But then what?: y' = 1/ (1 - ((2x+1)^2)

y = arccos(e^2x)

Other than plugging this into the formula for the derivative of arccos, I don't know what to do...

Also, is this implicit derivation problem? If so, why?:

ysin(x^2) = xsin(y^2)

product rule

ycosx^2 + y'sinx^2 = xcosy^2(2yy') +siny^2

y'sinx^2 - 2xyy'cosy^2 = -ycosx^2 +siny^2

ect, take out y' and solve...

2. Originally Posted by Maziana
My professor did not explain how to find the derivation of an inverse trig function very well, so I need some basic help in understanding what steps to take. I understand the derivations, and the need to use the chain rule and sometimes the product/quotient rule, but I don't understand these problems:

y = arcsin(2x +1)

Would this be the next step? But then what?: y' = 1/ (1 - ((2x+1)^2)

y = arccos(e^2x)

Other than plugging this into the formula for the derivative of arccos, I don't know what to do...

Also, is this implicit derivation problem? If so, why?:

ysin(x^2) = xsin(y^2)

product rule

ycosx^2 + y'sinx^2 = xcosy^2(2yy') +siny^2

y'sinx^2 - 2xyy'cosy^2 = -ycosx^2 +siny^2

ect, take out y' and solve...
$y = \arcsin(2x+1)$

$y' = \frac{2}{\sqrt{1 - (2x+1)^2}}$

$y = \arccos(e^x)$

$y' = -\frac{e^x}{\sqrt{1 - e^{2x}}}$

yes on implicit ...

$y\sin(x^2) = x\sin(y^2)$

$y \cdot 2x\cos(x^2) + \sin(x^2) \cdot y' = x \cdot 2y \cdot y' \cdot \cos(y^2) + \sin(y^2)$

solve for $y'$