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:

Remember \(\frac{d}{dx}\tan^{-1}x=\frac{1}{1+x^2}\).

Answer 1

The Correct Option is D

Step 1: Use formula: \[ \frac{d}{dx}(\tan^{-1}u)=\frac{1}{1+u^2}\frac{du}{dx} \]

Step 2: Here, \[ u=\frac{x}{a} \] \[ \frac{du}{dx}=\frac{1}{a} \]

Step 3: \[ \frac{d}{dx}\left(\tan^{-1}\frac{x}{a}\right) = \frac{1}{1+\left(\frac{x}{a}\right)^2}\cdot \frac{1}{a} \]

Step 4: \[ = \frac{1}{\frac{a^2+x^2}{a^2}}\cdot \frac{1}{a} \] \[ = \frac{a^2}{a^2+x^2}\cdot \frac{1}{a} \] \[ = \frac{a}{a^2+x^2} \] \[ \boxed{\frac{a}{a^2+x^2}} \]