# Differentiate the functions with respect to x in $\;\cos(\sqrt {x} \;)$

$\begin{array}{1 1} \large \frac{-sin \sqrt x}{2 \sqrt x} \\ \large \frac{cos \sqrt x}{2 \sqrt x} \\ \large \frac{-sin \sqrt x}{ \sqrt x} \\ \large \frac{cos \sqrt x}{ \sqrt x}\end{array}$

Toolbox:
• According to the Chain Rule for differentiation, given two functions $f(x)$ and $g(x)$, and $y=f(g(x)) \rightarrow y' = f'(g(x)).g'(x)$.
• $\; \large \frac{d\sqrt x}{dx}$$= \large \frac{1}{2 \sqrt x} • \; \large \frac{d(cosx)}{dx}$$=-sinx$
Given $y = \;\cos \sqrt {x}$. This is of the form $y = f(g(x))$ where $g(x) = \sqrt x$.
We can apply the Chain rule. According to the Chain Rule for differentiation, given two functions $f(x)$ and $g(x)$, and $y=f(g(x)) \rightarrow y' = f'(g(x)).g'(x)$.
$\; \large \frac{d\sqrt x}{dx}$$= \large \frac{1}{2 \sqrt x} \Rightarrow g'(x) = \large \frac{1}{2 \sqrt x} \; \large \frac{d(cosx)}{dx}$$=-sinx$
$\Rightarrow f'(g(x)) = -sin \sqrt x$
$\Rightarrow y' = f'(g(x)).g'(x) = -sin \sqrt x. \large \frac{1}{2 \sqrt x}$$= \large \frac{-sin \sqrt x}{2 \sqrt x}$