1. **Problem statement:** Given the function $y = \tan^{-1}\left(\frac{x}{a}\right)$, find the $n$th derivative $y^{(n)}$ with respect to $x$. 2. **Recall the formula for the first derivative:** $$y' = \frac{d}{dx} \tan^{-1}\left(\frac{x}{a}\right) = \frac{1}{1 + \left(\frac{x}{a}\right)^2} \cdot \frac{d}{dx}\left(\frac{x}{a}\right) = \frac{1}{1 + \frac{x^2}{a^2}} \cdot \frac{1}{a} = \frac{a}{a^2 + x^2}$$ 3. **Rewrite the first derivative:** $$y' = \frac{a}{a^2 + x^2}$$ 4. **General pattern for higher derivatives:** The $n$th derivative of $y$ with respect to $x$ can be found by differentiating $y'$ repeatedly. The derivatives of the function $f(x) = \frac{1}{a^2 + x^2}$ follow a known pattern involving factorials and powers. 5. **Using the formula for derivatives of $f(x) = (a^2 + x^2)^{-1}$:** $$\frac{d^n}{dx^n} (a^2 + x^2)^{-1} = (-1)^n \cdot \frac{(2n)!}{n!} \cdot \frac{x^n}{(a^2 + x^2)^{n+1}}$$ 6. **Apply the chain rule and factor $a$ from $y'$:** Since $y' = a \cdot (a^2 + x^2)^{-1}$, $$y^{(n)} = a \cdot \frac{d^n}{dx^n} (a^2 + x^2)^{-1} = a \cdot (-1)^n \cdot \frac{(2n)!}{n!} \cdot \frac{x^n}{(a^2 + x^2)^{n+1}}$$ 7. **Final expression for the $n$th derivative:** $$\boxed{y^{(n)} = a (-1)^n \frac{(2n)!}{n!} \frac{x^n}{(a^2 + x^2)^{n+1}}}$$ This formula gives the $n$th derivative of $y = \tan^{-1}\left(\frac{x}{a}\right)$ with respect to $x$.