Calculus Concepts: Inverse Functions, Inverse Trigonometry, and L'Hôpital's Rule

$$\frac{dF}{dx}(x) = \frac{1}{f'(f^{-1}(x))} = \frac{1}{f'(F(x))}$$

. This relationship tells us that the slope of the inverse function at a certain point is the reciprocal of the slope of the original function at the corresponding point.

Inverse Trigonometric Functions

Standard trigonometric functions like $\sin(x)$ are periodic, meaning they repeat values. To create an inverse, we must restrict their domains so they become "one-to-one". These are known as inverse trigonometric functions.

Common Inverse Trig Functions

The following table summarizes the primary inverse trigonometric functions, their domains, and their ranges:

Derivatives of Inverse Trig Functions

When differentiating these functions, we often use the chain rule. If $u$ is a function of $x$, the derivatives are:

• $\frac{d}{dx} \text{Arc}\sin(u) = \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}$
• $\frac{d}{dx} \text{Arc}\cos(u) = -\frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}$
• $\frac{d}{dx} \text{Arc}\tan(u) = \frac{1}{1+u^2} \cdot \frac{du}{dx}$
• $\frac{d}{dx} \text{Arc}\cot(u) = -\frac{1}{1+u^2} \cdot \frac{du}{dx}$

L’Hospital’s Rule and Indeterminate Forms

In many limit problems, direct substitution results in an indeterminate form, such as $\frac{0}{0}$ or $\frac{\infty}{\infty}$. These forms do not provide a definitive value for the limit and require further investigation.

L’Hospital’s Rule provides a method to resolve these limits by using derivatives.

The Rule

Let $f$ and $g$ be two functions defined and differentiable in a punctured neighborhood of $a$ (where $a$ is finite). If $g'(x) \neq 0$ in that neighborhood, and the limit $\lim_{x \to a} \frac{f(x)}{g(x)}$ results in the indeterminate forms $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then:

$$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$$

.

Pro Tip: This rule is not limited to finite values of $a$. It also holds if $a$ is replaced by $\pm\infty$ or if you are evaluating one-sided limits.

Example Application

Consider the limit: $\lim_{x \to 0} \frac{\sin(x)}{x}$

Direct substitution gives $\frac{\sin(0)}{0} = \frac{0}{0}$, which is indeterminate. Applying L’Hospital’s Rule: $\lim_{x \to 0} \frac{\frac{d}{dx}\sin(x)}{\frac{d}{dx}x} = \lim_{x \to 0} \frac{\cos(x)}{1} = \cos(0) = 1$

By differentiating the numerator and denominator separately, we successfully found the limit.