Question

$y = \sqrt{x} - \frac{1}{2 \sqrt{x} } + cosx$
find
$\frac{dy }{dx} =$
​

Answer 1

Question:

$\tt y = \sqrt{x} - \frac{1}{2 \sqrt{x} } + cosx \: \: find ~dy/dx$

Solution:

$\tt y = \sqrt{x} - \frac{1}{2 \sqrt{x} } + cosx \\\\ \to \tt \frac{dy} {dx} = d\frac{\bigg(\sqrt{x} - \frac{1}{2 \sqrt{x} } + cosx \bigg)}{dx} \\\\\to \tt d\frac{\sqrt{x}}{dx} - d \frac{ \frac{1}{2 \sqrt{x} }} {dx} + d \frac{cosx} {dx} \\\\\to \tt d \frac{x^{ \frac{1}{2}} }{dx} - \frac{1}{2}. \frac{d} {dx} \bigg[\frac{1}{\sqrt{x}} \bigg] +d \frac{cosx} {dx} \\\\\to \tt \frac{1}{2}x^{\frac{-1}{2}}- \frac{\big( - \frac{1}{2}x^ { - \frac{1}{2}-1} \big)} {2} - sinx \\\\ \to\tt \frac{1}{2 \sqrt{x}} + \frac{1}{4x^{\frac{3} {2}}} - sinx$