Question

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

• 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:

Use \(\frac{d}{dx}(\tan^{-1}u)=\frac{u'}{1+u^2}\). For \(u=\frac{x}{a}\), simplify carefully to get \(\frac{a}{a^2+x^2}\).

Answer 1

The Correct Option is D