Show that $\frac{d}{dx}\tan x=\sec^{2}x$ (by using infinitesimal approach to the derivative)

Question

I couldn't prove the derivative of $\tan x$ by using the infinitesimal approach to the derivatives. Here's my solution, but I stuck at the last step:

\begin{align} \tag{1}\frac{\mathrm{d} }{\mathrm{d} x}\tan x &= \frac{\tan (x+dx)- \tan x}{dx} \\ \tag{2} &=\frac{\frac{\tan x + \tan dx}{1-\tan x . \tan dx}-\tan x}{dx}\\ \tag{3}&=\frac{\frac{\tan x + dx - \tan x + dx\tan^2 x}{1 - dx\tan x}}{dx}\\ \tag{4} &=\frac{dx(1+\tan^2 x)}{1 - dx\tan x} \cdot \frac{1}{dx}=\frac{1+ \tan^2 x}{1-dx\tan x}\\ \tag{5} &=\frac{\sec^2 x}{1-dx\tan x}\neq \sec^2 x. \end{align}

Answer 1

You should not expect $\frac{1}{dx}(f(x + dx) - f(x))$ to be exactly $f'(x)$ because you would not expect this when $dx$ is a real number and whether $dx$ denotes a real number or an infinitesimal doesn't change the expression.

For instance

$$ \frac{(x + dx)^2 - x^2}{dx} = 2x + dx$$

whether or not you are thinking about $dx$ as a real number or an infinitesimal.

In standard analysis, we take the limit as $dx \to 0$. In non-standard analysis we take the standard part, which accomplishes much of the same thing.

The standard part of $2x + dx$ is $2x$. The standard part of

$$\frac{\sec^2 x}{1-dx\tan x} = \sec^2x(1 + \tan(x)dx + \tan(x)^2dx^2 + \tan(x)^3dx^3 + \cdots)$$

is $\sec^2x$.

Possibly there's a rule like $\operatorname{std}(a/b) = \operatorname{std}(a) / \operatorname{std}(b)$ you can use here too but I'm not 100% sure.