Let a function $y=y(x)$ be defined parametrically by $x(t)=2t-|t|, y(t)=t^2+t|t|, t \in R .$ Then $\frac{dy}{dx}$ at x= 5 isequal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 20
At $x=5$
$x(t)=2t-t$, $y(t)=t^2+t^2=2t^2$
$\frac{dx}{dt}=1$, $\frac{dy}{dt}=4t$
$⇒\frac{dy}{dx}=4t$
$⇒\left.\frac{dy}{dx}\right|_{t=5}=4×5=20$