Question

If $y={\tan }^{-1}\left(3x\right)$, then find $\frac{d^{2}y}{d{x}^2}$.

Answer 1

Given

$y=ta{n}^{-1}\left(3x\right)$

$\frac{dy}{dx}=\frac{d}{dx}ta{n}^{-1}\left(3x\right)$

$\frac{dy}{dx}=\frac{1}{1+(3x{)}^2}$

$\frac{dy}{dx}=\frac{1}{1+9x^2}$

$\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{1}{(1+9x^2)}\right)$

$\frac{d^{2}y}{d{x}^2}=\frac{(1+9x^2)\frac{d}{dx}3-3\frac{d}{dx}(1+9x^2)}{(1+9x^2{)}^2}$

$\frac{d^{2}y}{d{x}^2}=\frac{0-3\left(18x\right)}{(1+9x^2{)}^2}$

$\frac{d^{2}y}{d{x}^2}=\frac{-54x}{(1+9x^2{)}^2}$