- #1
Chocolaty
- 48
- 0
How do you derive a function where the trig is under a square root
like this:
y=sqr(tan^-1(x))
y=(tan^-1(x))^(1/2)
i know that y=tan^-1u(x) => dy/dx=1/(1+u^2)*du/dx
But how do I work it out? The book doesn't give the answers to pair numbers :(
like this:
y=sqr(tan^-1(x))
y=(tan^-1(x))^(1/2)
i know that y=tan^-1u(x) => dy/dx=1/(1+u^2)*du/dx
But how do I work it out? The book doesn't give the answers to pair numbers :(