Derivatives of Inverse Trig Functions — Question 5

Question 5

Let $$f(x) = \sec^{-1}\left( \frac{2x + 3}{x - 1} \right)$$

• Determine the domain of $f(x)$.

• Compute $f'(x)$.

Question 5 - Solution

(a) Domain:

The inverse secant function is defined for $|u| \ge 1$. Thus, we require $$\left| \frac{2x + 3}{x - 1} \right| \ge 1$$

Squaring both sides: $$\frac{(2x + 3)^2}{(x - 1)^2} \ge 1 \quad \Rightarrow \quad (2x + 3)^2 \ge (x - 1)^2$$

Expanding: $$4x^2 + 12x + 9 \ge x^2 - 2x + 1$$ $$3x^2 + 14x + 8 \ge 0$$

Factor: $$(3x + 2)(x + 4) \ge 0$$

Critical points are $x = -4$ and $x = -\tfrac{2}{3}$. Since the quadratic opens upward, the inequality holds outside the interval between the roots: $$x \le -4 \quad \text{or} \quad x \ge -\tfrac{2}{3}$$

We must also exclude $x = 1$, where the denominator is zero.

Therefore, the domain of $f(x)$ is: $$\boxed{ (-\infty,-4] \cup \left[-\tfrac{2}{3},1\right) \cup (1,\infty) }$$

(b) Derivative:

Using the derivative formula: $$\frac{d}{dx} \left( \sec^{-1}(u) \right) = \frac{1}{|u|\sqrt{u^2 - 1}} \cdot \frac{du}{dx}$$

Let $$u = \frac{2x + 3}{x - 1}$$

Then $$\frac{du}{dx} = \frac{(x - 1)(2) - (2x + 3)(1)}{(x - 1)^2} = \frac{-5}{(x - 1)^2}$$

Substitute into the formula: $$f'(x) = \frac{1}{\left| \frac{2x + 3}{x - 1} \right| \sqrt{\left( \frac{2x + 3}{x - 1} \right)^2 - 1}} \cdot \frac{-5}{(x - 1)^2}$$

$$\boxed{ f'(x) = \frac{-5}{ (x - 1)^2 \left| \frac{2x + 3}{x - 1} \right| \sqrt{\left( \frac{2x + 3}{x - 1} \right)^2 - 1} } }$$