Derivatives of Arctan and arcsin

[phypsychs]

I am confused about how to find arctan and arcsin

Specific Problem: y= arctan(4x/7) find derivative with respect to y

I know that d/dx arctan is 1/(1+x^2) am stuck on what to do. Any help would be awesome thanks!*p.s. i am very new to this site and it looks awesome! also not exactly sure where to post things so please any advice on posting would also be helpful

 

[matt_crouch]

if y=arctan(4x/7) what is the inverse of this?

 

[HallsofIvy]

matt crouch is suggesting that you find x as a function of t, then find dx/dy and "invert". That allows you to use tan instead of arctan.

But since you say you know the derivative of arctan so just use the chain rule.
Since d(arctan(u))/du= 1/(1+u²), d(arctan(u))/dx= (1/(1+u^2))du/dx where u= 4x/7.