Question

Differentiate both question ?

its urgent anyone know the answer

❤Don't spam the question ❤

Chapter is✌ Differentiation ✌

Hope u will help me in this question. ☺

Who know the answer then u post the answer if you not know the answer then don't try to answer plzz humbly request uuu....❤❤❤​

Answer 1

Step-by-step explanation:

Let

$y = \sqrt{ \frac{1 + \sin(x) }{1 - \sin(x) } } \: \: and \: \: z = \sqrt{ \cos(3x) }$

For y:

$y = \sqrt{ \frac{(1 + \sin(x) )(1 + \sin(x) )}{(1 - \sin(x))(1 + \sin(x) ) } }$

$= > y = \sqrt{ \frac{ {(1 + \sin(x)) }^{2} }{1 - \sin^{2} (x) } }$

$= > y = \frac{1 + \sin(x) }{ \cos(x) }$

$= > y = \sec(x) + \tan(x)$

Differentiating both sides, we get

$= > \frac{dy}{dx} = \sec(x) \tan(x) + \sec^{2} (x)$

$= > \frac{dy}{dx} = \frac{1 + \sin(x) }{ \cos^{2} (x) }$

For z:

$z = \sqrt{ \cos(3x) }$

Differentiating both sides, we get,

$= > \frac{dz}{dx} = \frac{1}{2 \sqrt{ \cos(3x) }} . \frac{d( \cos(3x)) }{dx}$

$= > \frac{dz}{dx} = - \frac{3 \sin(3x) }{2 \sqrt{ \cos(3x) } }$

Hope this will help you