Question

If $y=\tan x+secx$, prove that $\frac{d^{2}y}{d{x}^2}=\frac{\cos x}{(1-\sin x{)}^2}$.

Answer 1

Let $y=tanx+secx$.

To prove: $\frac{d^{2}y}{d{x}^2}=\frac{cosx}{(1-sinx{)}^2}$

$\frac{dy}{dx}=\frac{d}{dx}(tanx+secx)$

$=se{c}^{2}x+secxtanx$

$=\frac{1}{co{s}^{2}x}+\frac{1}{cosx}\cdot \frac{sinx}{cosx}$

$=\frac{1+sinx}{co{s}^{2}x}$

$=\frac{1+sinx}{1-si{n}^{2}x}$

$=\frac{1+sinx}{(1+sinx)(1-sinx)}$

$\therefore \frac{dy}{dx}=\frac{1}{1-sinx}=-\frac{1}{sinx-1}$

$\frac{d^{2}y}{d{x}^2}=-\frac{(sinx-1)\left(0\right)-\left(1\right)cosx}{(sinx-1{)}^2}$

$=\frac{cosx}{(sinx-1{)}^2}$

$\therefore \frac{d^{2}y}{d{x}^2}=\frac{cosx}{(1-sinx{)}^2}$