Question

If $y=\frac{\sec x-\tan x}{\sec x+\tan x}$, then $\frac{dy}{dx}$ equals?

• A. $2\sec x(\sec x-\tan x{)}^2$
• B. $-2\sec x(\sec x-\tan x{)}^2$
• C. $2\sec x(\sec x+tanx{)}^2$
• D. $-2\sec x(\sec x+\tan x{)}^2$

Answer 1

The correct option is B $-2\sec x(\sec x-\tan x{)}^2$

$y=\frac{secx-tanx}{secx+tanx}$

$\frac{dy}{dx}=\frac{(secx+tanx)(secxtanx-se{c}^{2}x)-(secx-tanx)(secxtanx+se{c}^{2}x)}{(secx+tanx{)}^2}$

$\frac{dy}{dx}=\frac{secx(secx+tanx)(tanx-secx)-secx(secx-tanx)(tanx+secx)}{(secx+tanx{)}^2}$

$\frac{dy}{dx}=\frac{2secx(tanx-secx)\times (tanx-secx)}{(tanx+secx)\times (tanx-secx)}$

$[ta{n}^{2}x-se{c}^{2}x-se{c}^{2}x=-1]$

$\frac{dy}{dx}=-2secx(secx-tanx{)}^2$