Derivative of $\arctan(x)$

Question

I am meant to find the derivative of $\arctan(x)$ from the definition of derivative of an inverse function $(1/(f '(f^{-1}(x)))$.

Well, I have found this question asked and explained, however all the explanations said that $(\tan(x))' = \sec^2(x)$.

I have never met with the $\sec$ function, so i have been trying to make it work with the formula $(\tan(x))' = 1/(\cos^2(x))$ instead. Thus writing this down:

$$\begin{align} ((f^{-1})(x)) &= \arctan (x)\\[1ex] (f(x)) &= \tan (x)\\[1ex] (f '(x)) &= \frac1{\cos^2 x} \end{align}$$

So from the inverse derivative formula the equation should look something like this:

$$(\arctan(x))' = \frac1{\frac1{\cos^2(\arctan x)}}$$

I rewrote the right side to

$$\frac{\sin^2(\arctan x)}{\tan^2(\arctan x)}$$

which equals

$$\frac{\sin^2(\arctan x)}{x^2}$$

And that is pretty much the furthest I can get, so if you could show me where I've done a mistake or how to go on with my solution, I'd be really thankful.

Answer 1

$$y=\tan x\iff x=\arctan y\implies (\arctan y)'=\frac1{\tan'\left(\arctan y\right)'}=\frac1{\frac1{\cos^2\arctan y}}=$$

$$=\cos^2\arctan y=\frac1{1+\tan^2\arctan y}=\frac1{1+y^2}$$

We used above the identity

$$1+\tan^2\alpha=\frac1{\cos^2\alpha}\iff \cos^2\alpha=\frac1{1+\tan^2\alpha}$$

Answer 2

Using the method you followed, you actually almost had found the answer. But you needed to be able to simplify the expression $\sin^2(\arctan(x)).$

Recall that $\arctan(x)$ is an angle $\theta$ such that $\tan(\theta) = x.$

Therefore $\sin^2(\arctan(x)) = \sin^2(\theta)$ where $\tan(\theta) = x.$

But if $\tan(\theta) = x,$ what is $\sin^2(\theta)$? This is something you can approach with algebra and trigonometric identities, but as a device to help me recall what the answer should be I find it useful sometimes to just construct a suitable right triangle. In this case the triangle has legs $1$ and $x$ so that the angle opposite leg $x$ is $\theta = \arctan(x).$

The hypotenuse of the triangle is $\sqrt{1+x^2},$ so looking at the angle opposite leg $x$, its sine (opposite/hypotenuse) is $$ \sin(\arctan(x)) = \sin(\theta) = \frac{x}{\sqrt{1+x^2}} $$ When we square this and divide by $x^2$ we have $$ \frac{\sin^2(\arctan(x))}{x^2} = \frac{\left(\frac{x}{\sqrt{1+x^2}}\right)^2}{x^2} = \frac{1}{1+x^2} $$