Question

find derivative of tan (root x) using first principle

Answer 1

here is your answer.

tan x is sec square x
root x is 1/2 root x

sec square /2 root x

I can't draw root ,square. so understand it.
hope this will help you

Answer 2

The derivative of  $f(x) = tan\sqrt{x}$ is given by

$f'(x) = \frac{1}{2\sqrt{x} } sec^{2}\sqrt{x}$

• Solving by 1st principle of derivative, we have

         $f'(x) = \lim_{h \to x\1} \frac{f(h)-f(x)}{h-x}$

         $f'(x) = \lim_{h \to x\1} \frac{tan\sqrt{h} -tan\sqrt{x} }{h-x}$

   Multiplying and dividing the above equation by $1+tan\sqrt{h} tan\sqrt{x}$ , we get

         $f'(x) = \lim_{h \to x\1} \frac{tan\sqrt{h} -tan\sqrt{x} }{1 + tan\sqrt{h}tan\sqrt{x} } (\frac{1+tan\sqrt{h}tan\sqrt{x} }{h-x})$

• Now, Using the identity

         $tan (A-B) = \frac{tanA - tanB}{1 + tanAtanB}$

      we get,

         $f'(x) = \lim_{h \to x\1} tan(\sqrt{h}-\sqrt{x} ) (\frac{1+tan\sqrt{h}tan\sqrt{x} }{h-x})$

         $f'(x) = \lim_{h \to x\1} \frac{tan(\sqrt{h} -\sqrt{x}) }{ h-x } \lim_{h \to x\1} 1+tan\sqrt{h}tan\sqrt{x}$

         $f'(x) = \lim_{h \to x\1} \frac{tan(\sqrt{h} -\sqrt{x}) }{( \sqrt{h}-\sqrt{x}) (\sqrt{h}+\sqrt{x} ) } ( 1+tan\sqrt{x}tan\sqrt{x})$

• Now, using the standard limits for tanx that is,

         $\lim_{x \to 0\1} \frac{tanx}{x} = 1$

         we get,

        $f'(x) = \frac{1 + tan^{2}\sqrt{x} }{\sqrt{x}+\sqrt{x} } \lim_{h \to x\1} \frac{tan(\sqrt{h} -\sqrt{x} )}{( \sqrt{h}-\sqrt{x}) }$

          $f'(x) = \frac{1 + tan^{2}\sqrt{x} }{2\sqrt{x} } (1)$

• Now, using identity

          $1 +tan^{2}x = sec^{2}x$, we get

          $f'(x) = \frac{sec^{2}\sqrt{x} }{2\sqrt{x} }$

   hence the derivative of f(x) by 1st principle is

               $f'(x) = \frac{sec^{2}\sqrt{x} }{2\sqrt{x} }$