Question

If $y=\log (1+2t^2+t^4),x={\tan }^{-1}t$, find $\frac{d^{2}y}{d{x}^2}$.

Answer 1

$y=\log (1+2t^2+t^4)$ $x={\tan }^{-1}t$

$y=\log (t^4+2t^2+1)$ $\frac{dx}{dt}=\frac{1}{1+t^2}$

$y=\log \left(\right(t^2+1{)}^2)$

$y=2\log (t^2+1)$

$\frac{dy}{dt}=\frac{2}{t^2+1}\left(2t\right)$

$\frac{dy}{dt}=\frac{4t}{t^2+1}$

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{\frac{4t}{t^2+1}}{\frac{1}{1+t^2}}=4t$

$\frac{d^{2}y}{d{x}^2}=\frac{4dt}{dx}=\frac{4}{\frac{dx}{dt}}=\frac{4}{\frac{1}{1+t^2}}=4(1+t^2)$

we have $x={\tan }^{-1}t$

$\Rightarrow \tan x=t$

$\therefore \frac{d^{2}y}{d{x}^2}=4(1+{\tan }^{2}x)=4{\sec }^{2}x$