Question

differentiate wrt x
sec x⁰ + tan x⁰ / sec x⁰ - tan x⁰

Answer 1

Hi,

$$\begin{gathered} \frac{secx+\tan x}{secx-\tan x} \\ =\frac{{\displaystyle secx+\tan x}}{{\displaystyle secx-\tan x}}\times \frac{{\displaystyle secx+\tan x}}{{\displaystyle secx+\tan x}} \\ =\frac{{\displaystyle {\left(secx+\tan x\right)}^2}}{{\displaystyle se{c}^{2}x-{\tan }^{2}x}} \\ ={\left(secx+\tan x\right)}^{2}asse{c}^{2}x-{\tan }^{2}x=1 \\ so\frac{dy}{dx}=2\left(secx+\tan x\right)\left(\left(secx\tan x+se{c}^{2}x\right)\right) \\ =2secx{\left(secx+\tan x\right)}^2 \end{gathered}$$
REGARDS