Question

Find the derivative using first principle : $\sin \sqrt x $

Answer 1

Hint: The derivative using the first principle is given by the formula $\mathop {\lim }\limits_{h \to 0} \left[ {\dfrac{{f(x + h) - f(x)}}{h}} \right]$ , where $f(x)$ is the function to be differentiation with respect to $x$ , and $h$ is tending to zero.

Complete step by step answer:
In the equation, we have to find the derivative of $\sin \sqrt x $ with respect to $x$ from first principles.
So, for that we will use the formula $\mathop {\lim }\limits_{h \to 0} \left[ {\dfrac{{f(x + h) - f(x)}}{h}} \right]$ , here the function to be differentiated is $f(x) = \sin \sqrt x $ . Now as per the formula, we have the required derivative as :
$ = \mathop {\lim }\limits_{h \to 0} \left[ {\dfrac{{f(x + h) - f(x)}}{h}} \right]$
$ = \mathop {\lim }\limits_{h \to 0} \left[ {\dfrac{{\sin \sqrt {(x + h)} - \sin \sqrt x }}{h}} \right]$
Here we can see that the limit is of the form $\dfrac{0}{0}$ , as we have
$\mathop {\lim }\limits_{h \to 0} \sin \sqrt {x + h} = \sin \sqrt x $
So $\mathop {\lim }\limits_{h \to 0} \left[ {\dfrac{{\sin \sqrt {(x + h)} - \sin \sqrt x }}{h}} \right] = \dfrac{0}{0}$
Hence , here we will use the L-Hospitals’ rule, where we differentiate the numerator and denominator separately, as follows:
$\mathop {\lim }\limits_{h \to 0} \left[ {\dfrac{{\sin \sqrt {(x + h)} - \sin \sqrt x }}{h}} \right]$
\[ = \mathop {\lim }\limits_{h \to 0} \left[ {\dfrac{{\dfrac{d}{{dh}}\left\{ {\sin \sqrt {(x + h)} - \sin \sqrt x } \right\}}}{{\dfrac{d}{{dh}}(h)}}} \right]\]
We know that the derivative of \[\dfrac{d}{{dh}}\left\{ {\sin \sqrt {(x + h)} } \right\} = \dfrac{{\cos \sqrt {x + h} }}{{2\sqrt {x + h} }}\] and \[\dfrac{d}{{dh}}\left\{ {\sin \sqrt x } \right\} = 0\]
Putting these in the above function and we get
$ = \mathop {\lim }\limits_{h \to 0} \left[ {\dfrac{{\left( {\cos \sqrt {(x + h)} } \right) \times \dfrac{1}{{2\sqrt {x + h} }} - 0}}{1}} \right]$
Simplifying, we get
$ = \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos \sqrt {x + h} }}{{2\sqrt {x + h} }}$
Putting $h = 0$ in the above function and we get
$ = \dfrac{{\cos \sqrt x }}{{2\sqrt x }}$
So final, we can say that
$\mathop {\lim }\limits_{h \to 0} \left[ {\dfrac{{f(x + h) - f(x)}}{h}} \right]$
$ = \mathop {\lim }\limits_{h \to 0} \left[ {\dfrac{{\sin \sqrt {(x + h)} - \sin \sqrt x }}{h}} \right]$
Simplifying , we get
$ = \dfrac{{\cos \sqrt x }}{{2\sqrt x }}$
Hence the derivative of $\sin \sqrt x $ is $\dfrac{{\cos \sqrt x }}{{2\sqrt x }}$ .

Note: The limit of the expression is to be found carefully, and applying the L-Hospitals’ rule, we will separately differentiate the numerator and the denominator and will not use the quotient rule of differentiation.