Question

Sin(ax + b) Derive it fast by chain rule and also Derive sec(tan)

Answer 1

(1)Let y = (ax + b)

First, we will find Derivative of Sin.

$Cos(ax+b).\frac{d}{dx}(ax+b)$      .....{sinx = cosx&cosx=-sinx}

$cos(ax+b).a$

(2)Mate, there is a correction in your ques it's tan√x not tan x.

Now,

Let y = (tan√x)

First, we will find Derivative of sec,

$sec(tan \sqrt{x} )tan(tan\sqrt{x} ).\frac{d}{dx}(tan\sqrt{x} )$

$sec(tan \sqrt{x} )tan(tan\sqrt{x} )*sec^{2} \sqrt{x} \frac{1}{2\sqrt{x} }$

Answer 2

$\huge\bf{Answer:-}$

Refer the attachment.