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

We need to differentiate: \[ \tan^{-1}\frac{x}{a}. \] Let \[ y=\tan^{-1}\frac{x}{a}. \] We know the formula: \[ \frac{d}{dx}\left(\tan^{-1}u\right)=\frac{1}{1+u^2}\frac{du}{dx}. \] Here, \[ u=\frac{x}{a}. \] So, \[ \frac{du}{dx}=\frac{1}{a}. \] Now apply the formula: \[ \frac{dy}{dx} = \frac{1}{1+\left(\frac{x}{a}\right)^2}\cdot \frac{1}{a}. \] Simplify the denominator: \[ 1+\left(\frac{x}{a}\right)^2 = 1+\frac{x^2}{a^2}. \] \[ =\frac{a^2+x^2}{a^2}. \] Therefore, \[ \frac{dy}{dx} = \frac{1}{\frac{a^2+x^2}{a^2}}\cdot \frac{1}{a}. \] \[ = \frac{a^2}{a^2+x^2}\cdot \frac{1}{a}. \] \[ = \frac{a}{a^2+x^2}. \] Hence, \[ \frac{d}{dx}\left(\tan^{-1}\frac{x}{a}\right)=\frac{a}{a^2+x^2}. \]