Differentiate the following function with respect to x.
$\cos \sqrt{x}$ .

Hint: In order to crack this problem, we need to know the chain rule of differentiation. It is shown as follows, $\dfrac{df\left( x \right)}{dx}=\dfrac{dg\left( h\left( x \right) \right)}{dh\left( x \right)}\times \dfrac{dh\left( x \right)}{dx}$ . Also, we need to know the individual differentiation formulas for each sun functions like $\dfrac{d\cos x}{dx}=-\sin x$ and $\dfrac{d\sqrt{x}}{dx}=\dfrac{1}{2\sqrt{x}}$ .

Complete step-by-step answer:
We aim to find the differentiation of this function with respect to x.
Let the function be named as $f\left( x \right)=\cos \sqrt{x}$ .
The above function is in the form $f\left( x \right)=g\left( h\left( x \right) \right)$
By comparing with the above form we can see that,
$h\left( x \right)=\sqrt{x}$ and $g\left( \sqrt{x} \right)=\cos \sqrt{x}$ …………..(i)
To find the differentiation we need to use the chain rule of differentiation as the functions cannot be separated as such.
The chain rule states as follows,
$\dfrac{df\left( x \right)}{dx}=\dfrac{dg\left( h\left( x \right) \right)}{dh\left( x \right)}\times \dfrac{dh\left( x \right)}{dx}................\left( ii \right)$
Therefore, from (i) and (ii) substituting the values, we get,
$\dfrac{df\left( x \right)}{dx}=\dfrac{d\cos \sqrt{x}}{d\sqrt{x}}\times \dfrac{d\sqrt{x}}{dx}$

The differentiation $\cos x$ is $\dfrac{d\cos x}{dx}=-\sin x..............(iii)$
And, the differentiation of $\sqrt{x}$ is $\dfrac{d\sqrt{x}}{dx}=\dfrac{1}{2\sqrt{x}}................(iv)$
Substituting the values from equation (iii) and (iv) we get,
$\dfrac{df\left( x \right)}{dx}=-\sin \left( \sqrt{x} \right)\times \dfrac{1}{2\sqrt{x}}$
Simplifying it further we get,
$\dfrac{df\left( x \right)}{dx}=\dfrac{-\sin \sqrt{x}}{2\sqrt{x}}$ .
Hence, this is the required solution.

Note: We need to remember the standard formulas of differentiation such as $\dfrac{d\cos x}{dx}=-\sin x$ . One more important point not to be missed is there is a negative sign for differentiation of $\cos x$ . This sign can easily be missed. Also, in the formula for chain rule, $\dfrac{df\left( x \right)}{dx}=\dfrac{dg\left( h\left( x \right) \right)}{dh\left( x \right)}\times \dfrac{dh\left( x \right)}{dx}$ , The first time is differentiated with respect to $h\left( x \right)$ and not only $x$ .