Question

Derivative, by first principletitle=" \tan( \sqrt{x } ) " alt=" \tan( \sqrt{x } ) " align="absmiddle" class="latex-formula">

Answer 1

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

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

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