Question

find y'. y=tan^-1 x/2

Answer 1

\[\large y=\arctan\left(\frac{x}{2}\right)\]

Are you allowed to use the definition of the derivative of arctan? Or does your teacher want you to do the steps to work through it?

Answer 2

\[\large \color{orangered}{y=\arctan x}\]\[\large \color{orangered}{y'=\frac{1}{1+x^2}}\]Are we allowed to use this? :)

Answer 3

we have not learned the dervivative of arctan yet. So im guessing he want us to work out but i like that formula better so we can use it

Answer 4

Hah ok let's use the formula, it will be rather quick :D
If you wanna go through the steps after that we can do that.

\[\large y=\arctan\left(\frac{x}{2}\right)\]
\[\large y'=\frac{1}{1+\left(\dfrac{x}{2}\right)^2}\left(\frac{x}{2}\right)'\]

Ok any questions about how I plugged that in?
Instead of x our argument is x/2.
So anywhere we would normally plug in x, we'll plug in x/2.

We also need to apply the chain rule.
The prime on the outside of that bracket lets us know that we still need to take the derivative of that part.

Answer 5

the derivative of x/2 is 1/2 right? and then what do i do after that?

Answer 6

Yes good good c: sorry got distracted lol

Answer 7

or is it 1-x/2

Answer 8

Then from there, just simplify if you want :D\[\large y'=\frac{\left(\dfrac{1}{2}\right)}{1+\left(\dfrac{x}{2}\right)^2}\]If we multiply the top and bottom by 4 we get,\[\large y'=\frac{2}{4+x^2}\]Which looks a little bit nicer.

Answer 9

1-x/2? :o

Answer 10

oh okay i got it. and sorry i sometimes over analyze stuff. lol

Answer 11

heh ^^