First Year Mathematics A (Differentiation & Linear Algebra) (1 Viewer)

matchalolz

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This is just mindlessly plugging into formulas which should be avoided at all costs for this case. matchalolz does not have a good idea of how to do this fundamental question so giving formulas without explanation/proof is worthless for the purposes of learning which is top priority in this case.

Whenever you make a claim, it is best to provide proof because the reader does not necessarily believe whatever you say unless it's trivial.
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
 

pikachu975

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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
Read up about the chain rule, it's fundamental to these types of questions.

E.g. dy/dx = dy/dt * dt/dx
 

sida1049

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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.
 
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