Question

Find the derivative of \[\tan \sqrt x \] with respect to x, using first principle.

Answer 1

Hint: We will first understand how we use the first principle to find derivatives of any function, say f(x). Then, we will use that concept for the function given to us and thus have the required answer.

Complete step-by-step solution:
Let us first understand what we mean by first principle and how to find the derivative of a function using the first principle.
The first principle states that:-
$ \Rightarrow f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}$
Now, as per the given question, we have with us:-
\[ \Rightarrow f(x) = \tan \sqrt x \] ………………..(1)
Replace x by x + h in the equation (1), we will then get:-
\[ \Rightarrow f(x + h) = \tan \sqrt {x + h} \] ………………..(2)
Now, we will put the equations (1) and (2) in the formula for first principle as states above:-
$ \Rightarrow f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\tan \sqrt {x + h} - \tan \sqrt x }}{h}$ ……………………..(3)
Now, we know that we have a formula given by the following expression:-
$ \Rightarrow \tan (x - y) = \dfrac{{\tan x - \tan y}}{{1 + \tan x.\tan y}}$
Replacing x by $\sqrt {x + h} $ and y by x in this, we will get the following expression:-
\[ \Rightarrow \tan (\sqrt {x + h} - \sqrt x ) = \dfrac{{\tan \sqrt {x + h} - \tan \sqrt x }}{{1 + \tan \sqrt {x + h} .\tan \sqrt x }}\]
Rearranging the terms to get the following expression:-
\[ \Rightarrow \tan \sqrt {x + h} - \tan \sqrt x = \tan (\sqrt {x + h} - \sqrt x )\left\{ {1 + \tan \sqrt {x + h} .\tan \sqrt x } \right\}\]
Putting the above derived expression in the equation (3), we will then obtain:-
$ \Rightarrow f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\tan (\sqrt {x + h} - \sqrt x )\left\{ {1 + \tan \sqrt {x + h} .\tan \sqrt x } \right\}}}{h}$
Now, we will multiply and divide the RHS by $\sqrt {x + h} - \sqrt x $ to obtain the following:-
$ \Rightarrow f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\tan \left( {\sqrt {x + h} - \sqrt x } \right)\left( {\sqrt {x + h} - \sqrt x } \right)\left\{ {1 + \tan \sqrt {x + h} .\tan \sqrt x } \right\}}}{{h\left( {\sqrt {x + h} - \sqrt x } \right)}}$
We can write this as follows:-
$ \Rightarrow f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\tan \left( {\sqrt {x + h} - \sqrt x } \right)}}{{\sqrt {x + h} - \sqrt x }} \times \mathop {\lim }\limits_{h \to 0} \dfrac{{\sqrt {x + h} - \sqrt x }}{h} \times \mathop {\lim }\limits_{h \to 0} \left( {1 + \tan \sqrt {x + h} .\tan \sqrt x } \right)$
Now, we will make use of the fact that: $\mathop {\lim }\limits_{\theta \to 0} \dfrac{{\tan \theta }}{\theta } = 1$. Using this in the above expression, we will get:-
$ \Rightarrow f'(x) = 1 \times \mathop {\lim }\limits_{h \to 0} \dfrac{{\sqrt {x + h} - \sqrt x }}{h} \times \mathop {\lim }\limits_{h \to 0} \left( {1 + \tan \sqrt {x + h} .\tan \sqrt x } \right)$
Now, we can see that the above expression contains $\mathop {\lim }\limits_{h \to 0} \left( {1 + \tan \sqrt {x + h} .\tan \sqrt x } \right)$ in which we can put h=0 without any problem, so we will then obtain:-
$ \Rightarrow f'(x) = \left( {1 + {{\tan }^2}\sqrt x } \right) \times \mathop {\lim }\limits_{h \to 0} \dfrac{{\sqrt {x + h} - \sqrt x }}{h}$
Multiplying and dividing the RHS by $\sqrt {x + h} + \sqrt x $ to obtain the following:-
$ \Rightarrow f'(x) = \left( {1 + {{\tan }^2}\sqrt x } \right) \times \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {\sqrt {x + h} - \sqrt x } \right)\left( {\sqrt {x + h} + \sqrt x } \right)}}{{h\left( {\sqrt {x + h} + \sqrt x } \right)}}$
Now, we will use the formula: $(a + b)(a - b) = {a^2} - {b^2}$ to obtain the following:-
$ \Rightarrow f'(x) = \left( {1 + {{\tan }^2}\sqrt x } \right) \times \mathop {\lim }\limits_{h \to 0} \dfrac{{x + h - x}}{{h\left( {\sqrt {x + h} + \sqrt x } \right)}}$
Simplifying the RHS to get:-
$ \Rightarrow f'(x) = \left( {1 + {{\tan }^2}\sqrt x } \right) \times \mathop {\lim }\limits_{h \to 0} \dfrac{1}{{\sqrt {x + h} + \sqrt x }}$
We also know that we have an identity: $1 + {\tan ^2}\theta = {\sec ^2}\theta $. Making use of this, we will get:-
$ \Rightarrow f'(x) = {\sec ^2}\sqrt x \times \dfrac{1}{{2\sqrt x }}$

$\therefore $ The required derivative of \[\tan \sqrt x \] is $\dfrac{{{{\sec }^2}\sqrt x }}{{2\sqrt x }}$.

Note: The students must note that we did multiply and divide the expression in the beginning with $\sqrt {x + h} - \sqrt x $. We cannot multiply and divide by anything unless we are sure that it cannot be equal to 0. So, here we already know that $h \to 0$ which means that h is not equal to 0. Therefore, it can never be equal to 0 and we have no problem with multiplying or dividing by it.
The students must also know that the derivative exists only if the limit mentioned in the first principle exists.