Question

find the derivative of y =secxtanx​

Answer 1

GIVEN :–

• A Function y = sec(x)tan(x)

TO FIND :–

• Value of dy/dx = ?

SOLUTION :–

• We know that –

$\\ \bf \implies \dfrac{d(u.v)}{dx} =u \dfrac{dv}{dx} + v\dfrac{du}{dx}$

• So that –

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

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

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

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

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

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

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

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

Answer 2

Step-by-step explanation:

GIVEN :–

• A Function y = sec(x)tan(x)

TO FIND :–

• Value of dy/dx = ?

SOLUTION :–

• We know that –

$\\ \bf \implies \dfrac{d(u.v)}{dx} =u \dfrac{dv}{dx} + v\dfrac{du}{dx}$

• So that –

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

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

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

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

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

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

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

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