Question

If \(x=\sinh^{-1}t+\log(t^{2}+1)\) and \(y=\tan^{-1}t+\log|t|\), then \(\frac{dy}{dx}=\)

• A. \(\frac{t^{2}+t+1}{2t+\sqrt{t^{4}+t^{2}}}\)
• B. \(\frac{t^{2}+t+1}{2t+\sqrt{t^{2}+1}}\)
• C. \(\frac{t^{2}+t+1}{2t^{2}+\sqrt{t^{4}+t^{2}}}\)
• D. \(\frac{t^{2}+t+1}{2+\sqrt{1+t^{2}}}\)

Hint:

For parametric curves, first calculate \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) separately and then use \(\frac{dy}{dx}=\frac{dy/dt}{dx/dt}\).

Answer 1

The Correct Option is C

Concept: For parametric equations \(x=f(t)\) and \(y=g(t)\), the derivative is given by \[ \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}. \]

Step 1: Find \(\frac{dx}{dt}\).
Given \[ x=\sinh^{-1}t+\log(t^2+1) \] Differentiating, \[ \frac{dx}{dt} = \frac{1}{\sqrt{1+t^2}} +\frac{2t}{1+t^2} \] Taking LCM, \[ \frac{dx}{dt} = \frac{\sqrt{1+t^2}+2t}{1+t^2} \]

Step 2: Find \(\frac{dy}{dt}\).
\[ y=\tan^{-1}t+\log|t| \] Differentiating, \[ \frac{dy}{dt} = \frac{1}{1+t^2} +\frac{1}{t} \] \[ = \frac{t+t^2+1}{t(1+t^2)} = \frac{t^2+t+1}{t(1+t^2)} \]

Step 3: Calculate \(\frac{dy}{dx}\).
\[ \frac{dy}{dx} = \frac{\frac{t^2+t+1}{t(1+t^2)}} {\frac{\sqrt{1+t^2}+2t}{1+t^2}} \] \[ = \frac{t^2+t+1} {t(\sqrt{1+t^2}+2t)} \] \[ = \frac{t^2+t+1} {t\sqrt{1+t^2}+2t^2} \] Since \[ t\sqrt{1+t^2} = \sqrt{t^4+t^2} \] we obtain \[ \frac{dy}{dx} = \frac{t^2+t+1} {2t^2+\sqrt{t^4+t^2}} \] center minipage0.55

\( \frac{dy}{dx} = \frac{t^2+t+1} {2t^2+\sqrt{t^4+t^2}} \) minipage center