# Thread: differentiation of trig functions

1. ## differentiation of trig functions

Hi, i'm struggling on this question and would appreciate some hints as to what I do next.
Q: Given that x=tan(y/2), prove that dy/dx=2/1+x^2

so far i have got this;

dx/dy=[sec^2(y/2)]/2 dy/dx= 2/[cos^2(y/2)] = 2/1-sin^2(y/2)

i think i need to change sin^2(y/2) to tan^2(y/2) which will give me the x^2, unfortunately i can't think how. presumably the minus sign i have will become a plus as you can't square x and get a negative answer.

2. Originally Posted by Tom G
Hi, i'm struggling on this question and would appreciate some hints as to what I do next.
Q: Given that x=tan(y/2), prove that dy/dx=2/1+x^2

so far i have got this;

dx/dy=[sec^2(y/2)]/2 dy/dx= 2/[cos^2(y/2)] = 2/1-sin^2(y/2)

i think i need to change sin^2(y/2) to tan^2(y/2) which will give me the x^2, unfortunately i can't think how. presumably the minus sign i have will become a plus as you can't square x and get a negative answer.
dx/dy=[sec^2(y/2)]/2 is good (there's NO dy/dx at the end btw). But forget about the 2/[cos^2(y/2)] = 2/1-sin^2(y/2) business .....

dx/dy=[sec^2(y/2)]/2 therefore dy/dx = 2/[sec^2(y/2)].

Now remember the standard identity tan^2(y/2) + 1 = sec^2(y/2).

But x=tan(y/2).

Therefore dy/dx = ....

3. Originally Posted by Tom G
Hi, i'm struggling on this question and would appreciate some hints as to what I do next.
Q: Given that x=tan(y/2), prove that dy/dx=2/1+x^2

so far i have got this;

dx/dy=[sec^2(y/2)]/2 dy/dx= 2/[cos^2(y/2)] = 2/1-sin^2(y/2)

i think i need to change sin^2(y/2) to tan^2(y/2) which will give me the x^2, unfortunately i can't think how. presumably the minus sign i have will become a plus as you can't square x and get a negative answer.
i'd approach the problem like this:

$x = \tan \frac y2$

$\Rightarrow \arctan x = \frac y2$

$\Rightarrow y = 2 \arctan x$

$\Rightarrow \frac {dy}{dx} = \frac 2{1 + x^2}$

as desired