Question

\(\frac{d}{dx} \left( \tan^{-1} \frac{x}{a} \right) =\)

• A. \(\frac{a}{a^2 - x^2}\)
• B. \(\frac{1}{a^2 + x^2}\)
• C. \(\frac{1}{a^2 - x^2}\)
• D. \(\frac{a}{a^2 + x^2}\)

Hint:

The derivative of \(\tan^{-1}\left(\frac{x}{a}\right)\) is a standard formula in calculus. It's highly recommended to memorize this result, \(\frac{a}{a^2+x^2}\), and the similar one for \(\sin^{-1}\left(\frac{x}{a}\right)\) to save time in exams.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We need to find the derivative of the inverse tangent function \(\tan^{-1}\left(\frac{x}{a}\right)\).

Step 2: Key Formula or Approach:
We will use the chain rule along with the standard derivative of the arctangent function.
Standard Derivative: \(\frac{d}{du}(\tan^{-1} u) = \frac{1}{1+u^2}\).
Chain Rule: If \(y = f(u)\) and \(u = g(x)\), then \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\).

Step 3: Detailed Explanation:
Let \(y = \tan^{-1}\left(\frac{x}{a}\right)\).
Let \(u = \frac{x}{a}\). Then \(y = \tan^{-1}(u)\).
First, find the derivative of \(y\) with respect to \(u\):
\[ \frac{dy}{du} = \frac{1}{1+u^2} \]
Next, 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:
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \left(\frac{1}{1+u^2}\right) \cdot \left(\frac{1}{a}\right) \]
Substitute \(u = \frac{x}{a}\) back into the expression:
\[ \frac{dy}{dx} = \left(\frac{1}{1 + \left(\frac{x}{a}\right)^2}\right) \cdot \left(\frac{1}{a}\right) \]
Simplify the expression:
\[ \frac{dy}{dx} = \left(\frac{1}{1 + \frac{x^2}{a^2}}\right) \cdot \left(\frac{1}{a}\right) = \left(\frac{1}{\frac{a^2+x^2}{a^2}}\right) \cdot \left(\frac{1}{a}\right) \]
\[ \frac{dy}{dx} = \left(\frac{a^2}{a^2+x^2}\right) \cdot \left(\frac{1}{a}\right) \]
\[ \frac{dy}{dx} = \frac{a}{a^2+x^2} \]

Step 4: Final Answer:
The derivative is \(\frac{a}{a^2+x^2}\).