Question

please help answer fast​

Answer 1

Answer:

The answer to your question is option D

Answer 2

Question :–

$\\ \bf \to If \: \: y = log( \sec x + \tan x) \: \: then \: \: \dfrac{dy}{dx} = ?$

ANSWER :–

GIVEN :–

$\\ \bf \longrightarrow \:\: y = log( \sec x + \tan x)$

TO FIND :–

$\\ \bf \longrightarrow \dfrac{dy}{dx} = ?$

SOLUTION :–

$\\ \bf \implies \:\: y = log( \sec x + \tan x)$

• Difference with respect to 'x' –

$\\ \bf \implies \dfrac{dy}{dx} = \left( \dfrac{1}{ \sec x + \tan x} \right)( \sec x. \tan x + { \sec}^{2}x)$

$\\ \bf \implies \dfrac{dy}{dx} = \left( \dfrac{1}{ \sec x + \tan x} \right)( \tan x +\sec x)( \sec x)$

$\\ \bf \implies \dfrac{dy}{dx} = \left( \cancel \dfrac{ \sec x + \tan x}{ \sec x + \tan x} \right)( \sec x)$

$\\ \implies \large{ \boxed{ \bf \dfrac{dy}{dx} = \sec x }}$

USED FORMULA :–

$\\ \bf \: (1) \: \dfrac{d \{ log(x) \}}{dx} = \dfrac{1}{x}$

$\\ \bf \: (2) \: \dfrac{d \{ \sec x \}}{dx} = \sec(x). \tan(x)$

$\\ \bf \: (3) \: \dfrac{d \{ \tan x \}}{dx} = \sec^{2} (x)$