Q:

# derivative, by first principle$$\tan( \sqrt{x } )$$

Accepted Solution

A:
$$\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}h$$

Employ a standard trick used in proving the chain rule:

$$\dfrac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\cdot\dfrac{\sqrt{x+h}-\sqrt x}h$$

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

$$\displaystyle\left(\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\right)\cdot\left(\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt x}h\right)$$

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

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

$$\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan\sqrt x}{\sqrt{x+h}-\sqrt x}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{y+h'-y}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{h'}$$

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

So we find that

$$\dfrac{\mathrm d\tan\sqrt x}{\mathrm dx}=\dfrac{\sec^2\sqrt x}{2\sqrt x}$$