Question

find dy/dx ; y = tan inverse { sqrt (a-b/a+b) tanx/2 }

Answer 1

\[\Large\rm y=\arctan\left(\frac{\sqrt{\frac{a-b}{a+b}}~\tan x}{2}\right)\]This...?

Answer 2

no its tan(x/2) not the whole thing divided by 2

Answer 3

\[\Large\rm y=\arctan\left(\sqrt{\frac{a-b}{a+b}}~\tan \frac{x}{2}\right)\]Oh ok :o

Answer 4

We can use chain rule and do it the long way...
But I think I remember a shortcut for this ... trying to rememberrrrrr

Answer 5

Well if we do it the normal way,
do you remember your arctan x derivative?

Answer 6

yes

Answer 7

\[\Large\rm y=\arctan \color{royalblue}{x},\qquad\to\qquad y'=\frac{1}{1+\color{royalblue}{x}^2}\]something like that, yah?

Answer 8

\[\Large\rm y=\arctan\color{royalblue}{\left(\sqrt{\frac{a-b}{a+b}}~\tan \frac{x}{2}\right)}\]
\[\Large\rm y'=\frac{1}{1+\color{royalblue}{\left(\sqrt{\frac{a-b}{a+b}}~\tan \frac{x}{2}\right)}^2}~~\frac{d}{dx}\color{royalblue}{\left(\sqrt{\frac{a-b}{a+b}}~\tan \frac{x}{2}\right)}\]

Answer 9

Oh boy this is gonna be a doozy lol.
Understand how I plugged that in similar to the way we did with arctan x?
We then have chain rule happening.
So we have to multiply by the derivative of the inner function.

Answer 10

i tried doing this using the chain rule..its way too long.. do you remember the shortcut yet ?

Answer 11

It shouldn't be `too long`.
A and B are constants yes?
So you won't have product rule or quotient rule or anything like that.

Answer 12

\[\Large\rm y'=\frac{1}{1+\color{royalblue}{\left(\sqrt{\frac{a-b}{a+b}}~\tan \frac{x}{2}\right)}^2}~\left(\sqrt{\frac{a-b}{a+b}}~\sec^2 \frac{x}{2}\right)\frac{d}{dx}\frac{x}{2}\]That big ugly chain just gives you that, right?
And then you just have to chain once more.

Answer 13

\[\Large\rm \frac{d}{dx}\tan x = \sec^2x\]

Answer 14

ok

Answer 15

Since this nasty thing is just a constant, let's replace it. \[\sqrt{\frac{a-b}{a+b}}=C\]Next let's look at the equation this way to save us some of the headache: \[\tan(y)=C \tan(\frac{x}{2})\]Then take the implicit derivative \[\sec^2 (y)y'=\frac{C}{2}\sec^2(\frac{x}{2})\]We can rearrange to get \[y'=\frac{C}{2}\sec^2(\frac{x}{2})\cos^2(y)\]Now use the above relation of tan(y) to draw a triangle. So what's cosine of y? It's adjacent over hyptenuse right? Then we square it. \[y'=\frac{Csec^2(\frac{x}{2})}{2(C^2\tan^2(\frac{x}{2})+1)}\] This probably simplifies more and then you can plug C back in. This isn't really that fun of a problem unfortunately.