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Derivative, by first principle

\tan( √(x ) )

User Moishie
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\displaystyle\lim_(h\to0)\frac{\tan√(x+h)-\tan x}h

Employ a standard trick used in proving the chain rule:


(\tan√(x+h)-\tan x)/(√(x+h)-\sqrt x)\cdot\frac{√(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)(\tan√(x+h)-\tan x)/(√(x+h)-\sqrt x)\right)\cdot\left(\lim_(h\to0)\frac{√(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
\frac1{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'=√(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)(\tan√(x+h)-\tan\sqrt x)/(√(x+h)-\sqrt x)=\lim_(h'\to0)(\tan(y+h')-\tan y)/(y+h'-y)=\lim_(h'\to0)(\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


(\mathrm d\tan\sqrt x)/(\mathrm dx)=(\sec^2\sqrt x)/(2\sqrt x)
User Salminnella
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