What is the second derivative of inverse tangent?

The answer to this question is ##1/(1+x^2)##. However, it's better to understand why it's that. So here I will explain why.

So we're trying to find what the derivative of ##arctanx## is. Let's put it into a function.

##y=arctanx##

The next step is to represent this differently. ##tany=x##

We can possibly implicitly differentiate this. We know that ##d/dx(tan(u(x)))=sec^2(u(x))*u'(x)##

So here we implicitly differentiate, ##sec^2(y)*y'=1## We solve for ##y'##, and we get: ##y'=1/sec^2(y)##

But, is this what we are looking for? Not really. It's nothing close to ##1/(1+x^2)##. We need more resources; fortunately, our trigonometric relations always come handy during derivatives and integrations.

So let's use the one from the pythagorean theorem and compute it this way: ##cos^2(y)+sin^2(y)=1## ##cos^2(y)/cos^2(y)+sin^2(y)/cos^2(y)=1/cos^2(y)## We use this so that we end up with the secant function. => ##1+tan^2(y)=sec^2(y)##

So, let's replace this into our solution. => ##y'=1/sec^2(y)=1/(1+tan^2(y))##

Although closer, it still does not look like our solution, but do you remember when we wrote the function differently? Isn't ##tany=x##?

So we would just have to replace it, and we get: ##d/dxarctan(x)=1/(1+x^2)##.

I hope that this was helpful, and now I challenge you to find ##d/dxsec^-1(x)##. (inverse function of ##secx## and not ##1/secx##)

The best way to succeed in Calculus is to really comprehend and not memorize.

If you have any question, feel free to ask :).