Question

How do you use the definition of derivative to find the derivative of $\dfrac{1}{{\sqrt x }}$ ?

Answer 1

Hint : Here, we are given the function with the variable $x$ . Let us say this is the function $f\left( x \right)$ .Now, we are asked to find the derivative of the given function by using the definition only. We know that the definition of derivative uses the concept of limit. Therefore, we will first discuss the definition and after that solve for the derivative of the given function.
Formula used:
Definition of the derivative is given as:
The derivative of the function $f\left( x \right)$ with respect to $x$ is the function $f'\left( x \right)$ and it is defined by the formula:
$f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$

Complete step by step solution:
We are given the function $f\left( x \right) = \dfrac{1}{{\sqrt x }}$ .
As per the definition of the derivative:
$f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \dfrac{{\dfrac{1}{{\sqrt {x + h} }} - \dfrac{1}{{\sqrt x }}}}{h}$
Now, we will have to solve this limit.
Let us first multiply both the numerator and denominator by $\dfrac{1}{{\sqrt {x + h} }} + \dfrac{1}{{\sqrt x }}$ .

$$\mathop {\lim }\limits_{h \to 0} \dfrac{{\dfrac{1}{{\sqrt {x + h} }} - \dfrac{1}{{\sqrt x }}}}{h} \\\ = \mathop {\lim }\limits_{h \to 0} \dfrac{{\dfrac{1}{{\sqrt {x + h} }} - \dfrac{1}{{\sqrt x }}}}{h} \times \dfrac{{\dfrac{1}{{\sqrt {x + h} }} + \dfrac{1}{{\sqrt x }}}}{{\dfrac{1}{{\sqrt {x + h} }} + \dfrac{1}{{\sqrt x }}}} \;$$

In the numerator, we can apply the formula for the difference of the perfect square that is: $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$ .

$$= \mathop {\lim }\limits_{h \to 0} \dfrac{{{{\left( {\dfrac{1}{{\sqrt {x + h} }}} \right)}^2} - {{\left( {\dfrac{1}{{\sqrt x }}} \right)}^2}}}{{h\left( {\dfrac{1}{{\sqrt {x + h} }} + \dfrac{1}{{\sqrt x }}} \right)}} \\\ = \mathop {\lim }\limits_{h \to 0} \dfrac{{\dfrac{1}{{x + h}} - \dfrac{1}{x}}}{{h\left( {\dfrac{{\sqrt x + \sqrt {x + h} }}{{\sqrt x \sqrt {x + h} }}} \right)}} \\\ = \mathop {\lim }\limits_{h \to 0} \dfrac{{\dfrac{{x - x - h}}{{x\left( {x + h} \right)}}}}{{h\left( {\dfrac{{\sqrt x + \sqrt {x + h} }}{{\sqrt x \sqrt {x + h} }}} \right)}} \\\ = \mathop {\lim }\limits_{h \to 0} \dfrac{{ - h}}{{hx\left( {x + h} \right)}} \times \dfrac{{\sqrt x \sqrt {x + h} }}{{\sqrt x + \sqrt {x + h} }} \\\ = \mathop {\lim }\limits_{h \to 0} \dfrac{{ - \sqrt x \sqrt {x + h} }}{{x\left( {x + h} \right)\left( {\sqrt x + \sqrt {x + h} } \right)}} \;$$

Now, we will put the value of $h = 0$ .

$$= \dfrac{{ - \sqrt x \sqrt {x + 0} }}{{x\left( {x + 0} \right)\left( {\sqrt x + \sqrt {x + 0} } \right)}} \\\ = \dfrac{{ - x}}{{{x^2}\left( {2\sqrt x } \right)}} \\\ = \dfrac{{ - 1}}{{2{x^{\dfrac{3}{2}}}}} \;$$

Thus, by using the definition of derivative, we can get the derivative of the function $\dfrac{1}{{\sqrt x }}$ as $\dfrac{{ - 1}}{{2{x^{\dfrac{3}{2}}}}}$ .
So, the correct answer is “ $\dfrac{1}{{\sqrt x }}$ as $\dfrac{{ - 1}}{{2{x^{\dfrac{3}{2}}}}}$ ”.

Note : While solving this type of question where we need to solve the limit, we cannot directly put the values such that the denominator becomes zero. This is because it will give us an infinite value. Therefore, in this question, instead of putting the value of $h = 0$ in the first step, we have first done some mathematical operation so that we can have a non-zero value for our denominator.