CUET Preparation Today
If $x=5\, t^3,y =7\, t^4 ,$ then $\frac{d^2y}{dx^2}$ is equal to : |
$\frac{28}{15}t$ $\frac{28}{15\, t^2}$ $\frac{14}{5\, t^2}$ $\frac{28}{225\, t^2 }$ |
$\frac{28}{225\, t^2 }$ |
The correct answer is Option (4) → $\frac{28}{225\, t^2 }$ $x=5t^3,y=7t^4$ $\frac{dx}{dt}=15t^2$, $\frac{dy}{dt}=28t^3$ $\frac{dy}{dx}=\frac{28t^3}{15t^2}=\frac{28t}{15}$ $\frac{d^2y}{dx^2}=\frac{28}{15}×\frac{dt}{dx}=\frac{28}{225\, t^2}$ |
