Question

If \[ x = \frac{9t^2}{1 + t^4}, \quad y = \frac{16t^2}{1 - t^4} \] then \[ \frac{dy}{dx} = \]

• A. \( \frac{16}{9} \left( \frac{1 - t^4}{1 + t^4} \right)^3 \)
• B. \( \frac{16(1 - t^4)}{9(1 + t^4)} \)
• C. \( \frac{9(1 - t^4)}{16(1 + t^4)} \)
• D. \( \frac{16}{9} \left( \frac{1 + t^4}{1 - t^4} \right)^3 \)

Hint:

When differentiating parametric equations, always use the chain rule: \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \] and simplify carefully.

Answer 1

The Correct Option is D

Step 1: Compute \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).
\[ \frac{dx}{dt} = \frac{d}{dt} \left( \frac{9t^2}{1 + t^4} \right) \] Using the quotient rule: \[ \frac{dx}{dt} = \frac{(18t(1 + t^4) - 9t^2(4t^3))}{(1 + t^4)^2} \] \[ = \frac{18t + 18t^5 - 36t^5}{(1 + t^4)^2} \] \[ = \frac{18t - 18t^5}{(1 + t^4)^2} \] Similarly, for \( y \): \[ \frac{dy}{dt} = \frac{d}{dt} \left( \frac{16t^2}{1 - t^4} \right) \] \[ = \frac{32t(1 - t^4) + 16t^2(4t^3)}{(1 - t^4)^2} \] \[ = \frac{32t - 32t^5 + 64t^5}{(1 - t^4)^2} \] \[ = \frac{32t + 32t^5}{(1 - t^4)^2} \] Step 2: Compute \( \frac{dy}{dx} \).
\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \] \[ = \frac{16}{9} \left( \frac{1 + t^4}{1 - t^4} \right)^3 \] Thus, the required derivative is: \[ \mathbf{\frac{16}{9} \left( \frac{1 + t^4}{1 - t^4} \right)^3} \]