Question

Find the derivative of $ \tan \sqrt x $ w.r.t. $ x $ using first principle.

Answer 1

Hint: To solve this problem, we can use the chain rule and derive it directly. But, we are asked to find the derivative by using the first principle which is the basic method to find the derivative. Here, we also need to use a trigonometric formula for determining the required derivative.
Formulas used:
First principle of derivatives states that the derivative of function $ f\left( x \right) $ is given as $ f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h} $
 $
  \tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}} \\
   \Rightarrow \tan A - \tan B = \tan \left( {A - B} \right)\left( {1 + \tan A\tan B} \right) \;
  $

Complete step-by-step answer:
We have $ f\left( x \right) = \tan \sqrt x $ , therefore $ f\left( {x + h} \right) = \tan \sqrt {x + h} $
Now, we know that by the first principle, the derivative can is given by the following formula:
 $ f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h} $
Now, we will put $ f\left( x \right) = \tan \sqrt x $ and $ f\left( {x + h} \right) = \tan \sqrt {x + h} $ in this formula.
 $ \Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\tan \sqrt {x + h} - \tan \sqrt x }}{h} $
We know that $ \tan A - \tan B = \tan \left( {A - B} \right)\left( {1 + \tan A\tan B} \right) $ . We will use this formula for $ \tan \sqrt {x + h} - \tan \sqrt x $ .
We will put $ \tan \sqrt {x + h} - \tan \sqrt x = \tan \left( {\sqrt {x + h} - \sqrt x } \right)\left( {1 + \tan \sqrt {x + h} \tan \sqrt x } \right) $
 $ \Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\tan \left( {\sqrt {x + h} - \sqrt x } \right)\left( {1 + \tan \sqrt {x + h} \tan \sqrt x } \right)}}{h} $
Now we will separate the terms
 $ \Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\tan \left( {\sqrt {x + h} - \sqrt x } \right)}}{h}\mathop {\lim }\limits_{h \to 0} \left( {1 + \tan \sqrt {x + h} \tan \sqrt x } \right) $
We will find the limits separately first and put them into the main equation.
First, we will solve $ \mathop {\lim }\limits_{h \to 0} \dfrac{{\tan \left( {\sqrt {x + h} - \sqrt x } \right)}}{h} $
We can write $ h = \left( {\sqrt {x + h} - \sqrt x } \right)\left( {\sqrt {x + h} + \sqrt x } \right) $
 $
  \mathop {\lim }\limits_{h \to 0} \dfrac{{\tan \left( {\sqrt {x + h} - \sqrt x } \right)}}{h} \\
   = \mathop {\lim }\limits_{h \to 0} \dfrac{{\tan \left( {\sqrt {x + h} - \sqrt x } \right)}}{{\left( {\sqrt {x + h} - \sqrt x } \right)\left( {\sqrt {x + h} + \sqrt x } \right)}} \\
   = \mathop {\lim }\limits_{h \to 0} \dfrac{{\tan \left( {\sqrt {x + h} - \sqrt x } \right)}}{{\left( {\sqrt {x + h} - \sqrt x } \right)}}\mathop {\lim }\limits_{h \to 0} \dfrac{1}{{\left( {\sqrt {x + h} + \sqrt x } \right)}} \\
   = 1 \times \dfrac{1}{{2\sqrt x }} \\
   = \dfrac{1}{{2\sqrt x }} \;
  $
Now, we will solve the second term \[\mathop {\lim }\limits_{h \to 0} \left( {1 + \tan \sqrt {x + h} \tan \sqrt x } \right)\]
\[
  \mathop {\lim }\limits_{h \to 0} \left( {1 + \tan \sqrt {x + h} \tan \sqrt x } \right) \\
   = \left( {1 + \tan \sqrt x \tan \sqrt x } \right) \\
   = 1 + {\tan ^2}\sqrt x \\
   = {\sec ^2}\sqrt x \;
 \]
Now, we will put both these values in the main equation
 $
   \Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\tan \left( {\sqrt {x + h} - \sqrt x } \right)}}{h}\mathop {\lim }\limits_{h \to 0} \left( {1 + \tan \sqrt {x + h} \tan \sqrt x } \right) \\
   \Rightarrow f'\left( x \right) = \dfrac{1}{{2\sqrt x }}{\sec ^2}\sqrt x \;
  $
Thus, the derivative of $ \tan \sqrt x $ w.r.t. $ x $ using the first principle is $ \dfrac{1}{{2\sqrt x }}{\sec ^2}\sqrt x $ .
So, the correct answer is “$ \dfrac{1}{{2\sqrt x }}{\sec ^2}\sqrt x $ .”.

Note: There is a second method to find the derivative of $ \tan \sqrt x $ w.r.t. $ x $ . This method is called the chain rule. In this method we will find the derivative of the main function which is whose derivative is $ {\sec ^2}y $ . Here, we have taken $ \sqrt x = y $ . Now, we will find the derivative of $ \sqrt x $ which is $ \dfrac{1}{{2\sqrt x }} $ . Now, we will multiply both these derivatives and get the final answer as $ \dfrac{1}{{2\sqrt x }}{\sec ^2}\sqrt x $ . This can also be written as:
 $ \dfrac{d}{{dx}}\tan \sqrt x = \dfrac{1}{{2\sqrt x }}{\sec ^2}\sqrt x $