Yeah pretty much I don't even have the basics down

I think I'll probably ask someone in real life because I'm so clueless rip

To be completely honest, most HSC students who come across similar questions apply the chain rule along with the formula of the differentiation of inverse tan, without knowing where central formula actually comes from. You should probably see it at least once...

I'll offer an elementary derivation of the differentiation of inverse tan formula. I'll denote inverse tan as

*arctan*.

Okay, so we want to show that

*d(arctan(x))/dx = 1/(1+x^2)*.

First, let

*y = arctan(x)*. Since the inverse tan is, well, an inverse, we have that

*x = tan(y)*.

Differentiating with respect to

*y*, we have that

*dx/dy = dtan(y)/dy = sec^2(y)*. The last equality (that differentiating

*tan(y)* gives you

*sec^2(y)*) is also something that HSC students normally memorise. If you want to

*see* it, take

*tan(y) = sin(y)/cos(y)* and apply the quotient rule.

Now you need to note that

*sin^2(y) + cos^2(y) = 1*. This is a famous identity that can be proven in quite a lot of ways (for example, considering aright-angled triangle made by a radius of a unit circle and the axes, and applying Pythagoras' Theorem). If we divide both sides by

*cos^2(y)*, you get

*tan^2(y) + 1 = sec^2(y)*. Substituting this into

*dx/dy = sec^2(y)*, we get that

*dx/dy = 1 + tan^2(y)*.

Now remember that $x = tan^2(y)$, so

*dx/dy = 1 + x^2*. And so finally, by taking the reciprocal, we get the formula that

*darctan(y)/dx =1/(1 + x^2)*.

Now in your problem, we need to replace our

*y* with

*2/y*. This complicates things a bit, because now we need to apply the chain rule, which pikachu975 has stated as

*dz/dx = dz/dt * dt/dx*, where

*z = arctan(t)* and

*t = 2/y*.

So

*dz/dx = darctan(t)/dt * d(2/y)/dy*.

From here, apply the inverse tan differentiation we have derived to the first term (you get

*1/(1+t^2)*), and differentiate the second term (you get

*-2/y^2*). Then substitute $t=2/y$ back in to get your answer.

Hope this helps.