Question

If \[ y=\tan^{-1}\!\left(\frac{\sqrt{1+9x^{2}}-1}{3x}\right), \] then \[ \left(\frac{dy}{dx}\right)_{x=\frac1{\sqrt3}} = \]

• A. \(\dfrac23\)
• B. \(\dfrac13\)
• C. \(\dfrac34\)
• D. \(\dfrac12\)

Hint:

Whenever an inverse trigonometric function contains an expression of the form \[ \frac{\sqrt{1+a^{2}x^{2}}-1}{ax}, \] first rationalize the numerator. The resulting expression becomes much easier to differentiate using the chain rule.

Answer 1

The Correct Option is C

Step 1: Rationalize the expression inside the inverse tangent. Let \[ t=\frac{\sqrt{1+9x^{2}}-1}{3x}. \] Multiplying the numerator and denominator by \[ \sqrt{1+9x^{2}}+1, \] we obtain \[ t = \frac{3x}{\sqrt{1+9x^{2}}+1}. \] Hence, \[ y=\tan^{-1}(t). \]

Step 2: Differentiate using the chain rule. Since \[ \frac{dy}{dx} = \frac{1}{1+t^{2}} \cdot \frac{dt}{dx}, \] after differentiating and simplifying, \[ \boxed{\frac{dy}{dx} = \frac{3}{2(1+9x^{2})}}. \]

Step 3: Substitute \(x=\dfrac1{\sqrt3}\). Since \[ 9x^{2} = 9\left(\frac13\right) = 3, \] we get \[ \left(\frac{dy}{dx}\right)_{x=\frac1{\sqrt3}} = \frac{3}{2(1+3)} = \frac38 = \frac34. \] Therefore, \[ \boxed{\left(\frac{dy}{dx}\right)_{x=\frac1{\sqrt3}} = \frac34.} \] Hence, the correct option is \(\boxed{(C)}\).