CUET Preparation Today
CUET
-- Mathematics - Section A
Continuity and Differentiability
If $x=t^2$ and $y=t^3$, then $\frac{d y}{d x}=6$ at t =
1
3
4
6
The correct answer is Option (3) → 4
$\frac{dx}{dt}=2t$
$\frac{dy}{dt}=3t^2$
so $\frac{dy}{dx}=\frac{3}{2}t$
if $\frac{3}{2}t=6⇒t=4$
