CUET Preparation Today
If $x=\frac{1-t}{1+t} $ and $y = 2t^3 $. then, value of $\frac{dy}{dx} $ at $t =1$ is : |
-16 -12 6 11 |
-12 |
The correct answer is Option (2) → -12 $\frac{dy}{dt}=6t^2$ and $\frac{dx}{dt}=\frac{-1(1+t)+(1-t)}{(1+t)^2}$ $\frac{dy}{dx}=\frac{6t^2×(1+t)^2}{-2t}$ $\left.\frac{dy}{dx}\right|_1=\frac{6×4}{-2}$ $=-12$ |
