Is the Primer Introduction to Derivation of Implicit Functions Using This Definition Ambiguous?

Question

I'm exploring the concept of implicit function derivation and parameterized curve derivation, and I've encountered some points that seem vague to me. I would appreciate any insights or clarifications on these topics.

Simple Derivation of Implicit Functions and Parameterized Curves:

• The introduction I've seen using the chain rule and applying the derivative of the inverse function in a manner that seems somewhat imprecise. Specifically, the requirement for the derivative of the inverse function, $ f'(x_0) \ne 0 $, originates from the denominator being a constant 1. However, in this context, we first obtain the ratio of the derivatives of two parametric equations and then take the limit, which might lead to an indeterminate form.
• For example, consider the equation: $$ \left\{ \begin{align*} y(t) &= t - \ln(1 + t) \\ x(t) &= t^3 + t^2 \end{align*} \right. $$
• Here, $x'(t) = 3t^2 + 2t$, meaning $x'(0) = 0$ . However, $y'(0)=x'(0)=0$, and we don't even need to apply L'Hôpital's Rule, since $\frac{y'(t)}{x'(t)} = \frac{\frac{t}{1 + t}}{3t^2 + 2t} = \frac{1}{(1 + t)(3t + 2t)}$ . So, if we take $\mathop{\lim} \limits _{t \rightarrow 0} \frac{y'(t)}{x'(t)}$ (the derivative is defined as a limit), it seems to equal $\frac{1}{2}$ simply.

Reference:

Derivation of Implicit Functions and Parameterized Curves:

• The derivative of $ y(t) $ with respect to $ x(t) $ at a specific point $ x_0 = x(t_0) $ ($x'(t_0) \ne 0$) can be expressed as follows:
  • $\frac{dy(t)}{dx(t)}|_{x_0} = \frac{dy(t)}{dt}|_{t_0} \frac{dt}{dx(t)}|_{x_0} = \frac{\frac{dy(t)}{dt}|_{t_0}}{\frac{dx(t)}{dt}|_{t_0}} = \frac{y'(t_0)}{x'(t_0)}$
• The first equality is based on the chain rule, and the second equality involves applying the derivative of the inverse function, which requires the condition "$x'(t_0) \ne 0$".

Theorem - Derivative and Differential of the Inverse Function:

• Let $ X, Y \subset \mathbb{R} $, with $ x_0 \in X $ as a limit point and $ y_0 \in Y $ also as a limit point. Suppose $ f: X \mapsto Y $ and $ f^{-1}: Y \mapsto X $ with $ f(x_0) = y_0 $ and $ x_0 = f^{-1} (y_0) $.
• If:
  • (i) $ f $ is differentiable at $ x_0 $ with $ f'(x_0) \ne 0 $;
  • (ii) $ f^{-1} $ is continuous at $ y_0 $;
• Then $ f^{-1} $ is differentiable at $ y_0 $, and $$ (f^{-1})'(y_0) = \frac{1}{f'(x_0)} = \frac{1}{f'(f^{-1}(y_0))} $$