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?

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' $