Step 1: Understanding the Concept:
This question asks for the derivative of the inverse tangent function with an argument that is a function of x. This requires applying the standard derivative formula for $\tan^{-1}(u)$ along with the chain rule.
Step 2: Key Formula or Approach:
The standard formula for the derivative of the inverse tangent function is:
\[ \frac{d}{du} (\tan^{-1}u) = \frac{1}{1+u^2} \]
When the argument $u$ is a function of $x$, we use the chain rule:
\[ \frac{d}{dx} (\tan^{-1}u) = \frac{1}{1+u^2} \cdot \frac{du}{dx} \]
Step 3: Detailed Explanation:
Let $y = \tan^{-1}\left(\frac{x}{a}\right)$.
Here, the argument is $u = \frac{x}{a}$.
First, find the derivative of $u$ with respect to $x$:
\[ \frac{du}{dx} = \frac{d}{dx}\left(\frac{x}{a}\right) = \frac{1}{a} \]
Now apply the chain rule formula:
\[ \frac{dy}{dx} = \frac{1}{1+u^2} \cdot \frac{du}{dx} \]
Substitute $u = \frac{x}{a}$ and $\frac{du}{dx} = \frac{1}{a}$:
\[ \frac{dy}{dx} = \frac{1}{1 + \left(\frac{x}{a}\right)^2} \cdot \frac{1}{a} \]
\[ \frac{dy}{dx} = \frac{1}{1 + \frac{x^2}{a^2}} \cdot \frac{1}{a} \]
To simplify the fraction, find a common denominator in the first term:
\[ \frac{dy}{dx} = \frac{1}{\frac{a^2+x^2}{a^2}} \cdot \frac{1}{a} \]
\[ \frac{dy}{dx} = \frac{a^2}{a^2+x^2} \cdot \frac{1}{a} \]
Cancel one 'a' from the numerator and denominator:
\[ \frac{dy}{dx} = \frac{a}{a^2+x^2} \]
Step 4: Final Answer:
The derivative of $\tan^{-1}\left(\frac{x}{a}\right)$ is $\frac{a}{a^2+x^2}$. Therefore, option (D) is correct.