Question

Diffrentiate y=

$\sqrt{x + \sqrt{x} }$
​

Answer 1

$\sqrt{x + \sqrt{x} }$

$= {x}^{ 1 \div 2} + {x}^{ 1 \div 4}$

on differentiating

$0.5 {x}^{ - 0.5} +0.25 {x}^{ - 3 \div 4}$

Answer 2

Answer:

Step-by-step explanation: