derivative, by first principle[tex] \tan( \sqrt{x } ) [/tex]

[tex]\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}h[/tex]

Employ a standard trick used in proving the chain rule:

[tex]\dfrac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\cdot\dfrac{\sqrt{x+h}-\sqrt x}h[/tex]

The limit of a product is the product of limits, i.e. we can write

[tex]\displaystyle\left(\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\right)\cdot\left(\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt x}h\right)[/tex]

The rightmost limit is an exercise in differentiating [tex]\sqrt x[/tex] using the definition, which you probably already know is [tex]\dfrac1{2\sqrt x}[/tex].

For the leftmost limit, we make a substitution [tex]y=\sqrt x[/tex]. Now, if we make a slight change to [tex]x[/tex] by adding a small number [tex]h[/tex], this propagates a similar small change in [tex]y[/tex] that we'll call [tex]h'[/tex], so that we can set [tex]y+h'=\sqrt{x+h}[/tex]. Then as [tex]h\to0[/tex], we see that it's also the case that [tex]h'\to0[/tex] (since we fix [tex]y=\sqrt x[/tex]). So we can write the remaining limit as

[tex]\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan\sqrt x}{\sqrt{x+h}-\sqrt x}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{y+h'-y}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{h'}[/tex]

which in turn is the derivative of [tex]\tan y[/tex], another limit you probably already know how to compute. We'd end up with [tex]\sec^2y[/tex], or [tex]\sec^2\sqrt x[/tex].

So we find that

[tex]\dfrac{\mathrm d\tan\sqrt x}{\mathrm dx}=\dfrac{\sec^2\sqrt x}{2\sqrt x}[/tex]