# Derivatives of inverse trig function

• Nov 11th 2011, 03:11 PM
Barthayn
Derivatives of inverse trig function
I need help taking the derivative of y = arctan(x+1)
This is what I do
y = arctan(x+1)
x=tan(y+1)
cos^2(y+1) = y'
cos^2(arctan(x+1)+1) = y'
cos^2(((1-sec^2x)^0.5)+2))=y'

I am suppose to get a polynomial but I cannot take the derivative of it to get one. I just get stuck at the last step. Any help?
• Nov 11th 2011, 03:14 PM
e^(i*pi)
Re: Derivatives of inverse trig function
Quote:

Originally Posted by Barthayn
I need help taking the derivative of y = arctan(x+1)
This is what I do
y = arctan(x+1)
x=tan(y+1)
cos^2(y+1) = y'
cos^2(arctan(x+1)+1) = y'
cos^2(((1-sec^2x)^0.5)+2))=y'

I am suppose to get a polynomial but I cannot take the derivative of it to get one. I just get stuck at the last step. Any help?

The tan and arctan functions don't work that way. You can only take the tangent of the whole expression

\$\displaystyle \tan(y) = \tan(\arctan(x+1))\$

\$\displaystyle \tan(y) = x+1\$

\$\displaystyle x = \tan(y) - 1\$

When you differentiate wrt x you should get: \$\displaystyle 1 = \sec^2(y)y' \$