Question Statement

Find the derivative of $y=\frac{{\sec }^{-1}x}{x}$ with respect to $x$.

---

Background and Explanation

This problem involves differentiating a quotient where the numerator is an inverse trigonometric function. You will need to apply the quotient rule along with the standard derivative formula for ${\sec }^{-1}x$.

---

Solution

We begin with the function:

$y=\frac{{\sec }^{-1}x}{x}$

This is a quotient of two functions: $u={\sec }^{-1}x$ (numerator) and $v=x$ (denominator). We apply the quotient rule, which states that for $y=\frac{u}{v}$:

$\frac{dy}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$

First, identify the individual derivatives:

• $\frac{d}{dx}({\sec }^{-1}x)=\frac{1}{x\sqrt{x^2-1}}$ (standard formula for the derivative of inverse secant, valid for $|x|>1$)
• $\frac{d}{dx}(x)=1$

Applying the quotient rule by substituting $u$, $v$, and their derivatives:

$\frac{dy}{dx}=\frac{x\cdot \frac{d}{dx}({\sec }^{-1}x)-{\sec }^{-1}x\cdot \frac{d}{dx}(x)}{x^2}$

Substitute the specific derivative values into the expression:

$\frac{dy}{dx}=\frac{x\cdot \frac{1}{x\sqrt{x^2-1}}-{\sec }^{-1}x\cdot 1}{x^2}$

Simplify the numerator by canceling the $x$ in the first term:

$\frac{dy}{dx}=\frac{\frac{1}{\sqrt{x^2-1}}-{\sec }^{-1}x}{x^2}$

Finally, factor out $\frac{1}{x^2}$ to write the answer in a compact form:

$\frac{dy}{dx}=\frac{1}{x^2}\left[\frac{1}{\sqrt{x^2-1}}-{\sec }^{-1}x\right]$

---

Key Formulas or Methods Used

• Quotient Rule: $\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$
• Derivative of Inverse Secant: $\frac{d}{dx}({\sec }^{-1}x)=\frac{1}{x\sqrt{x^2-1}}$ for $|x|>1$
• Power Rule: $\frac{d}{dx}(x)=1$

---

Summary of Steps

1. Identify the function as a quotient: $y=\frac{u}{v}$ where $u={\sec }^{-1}x$ and $v=x$
2. Recall the quotient rule formula and the derivative $\frac{d}{dx}({\sec }^{-1}x)=\frac{1}{x\sqrt{x^2-1}}$
3. Apply the quotient rule: $\frac{dy}{dx}=\frac{x\cdot \frac{1}{x\sqrt{x^2-1}}-{\sec }^{-1}x\cdot 1}{x^2}$
4. Simplify the numerator by canceling $x$ in the first term
5. Factor out $\frac{1}{x^2}$ to obtain the final result: $\frac{1}{x^2}\left[\frac{1}{\sqrt{x^2-1}}-{\sec }^{-1}x\right]$