Question

Using the limit definition, how do you find the derivative of $f(x) = \sqrt {2x - 1}$?

Answer 1

Hint : Here we need to find the derivative of the given function using the limit definition. We have $f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}$ . Here we have $f(x) = \sqrt {2x - 1}$ by substituting this In the formula we will get the desired answer.

Complete step by step solution:
Given,
$f(x) = \sqrt {2x - 1}$ .
From the definition of the derivatives we have,s
$f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}$ .
We need $f(x + h)$ , now we find $f(x + h)$ .
$f(x + h) = \sqrt {2(x + h) - 1} = \sqrt {2x + 2h - 1}$
Now substituting we have,
$\Rightarrow f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}$
$f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {\sqrt {2x + 2h - 1} } \right) - \left( {\sqrt {2x - 1} } \right)}}{h}$ .
If we apply the limit now we will have indeterminate method so we multiply and divide by
$\Rightarrow \left( {\sqrt {2x + 2h - 1} } \right) + \left( {\sqrt {2x - 1} } \right)$ for further simplification,
$\Rightarrow f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {\sqrt {2x + 2h - 1} } \right) - \left( {\sqrt {2x - 1} } \right)}}{h}\dfrac{{\left( {\sqrt {2x + 2h - 1} } \right) + \left( {\sqrt {2x - 1} } \right)}}{{\left( {\sqrt {2x + 2h - 1} } \right) + \left( {\sqrt {2x - 1} } \right)}}$
$\Rightarrow f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {\sqrt {2x + 2h - 1} } \right) - \left( {\sqrt {2x - 1} } \right)\left( {\sqrt {2x + 2h - 1} } \right) + \left( {\sqrt {2x - 1} } \right)}}{{h\left( {\sqrt {2x + 2h - 1} } \right) + \left( {\sqrt {2x - 1} } \right)}}$
The numerator is of the form $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$ then we will have,
$\Rightarrow f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{{{\left( {\sqrt {2x + 2h - 1} } \right)}^2} - {{\left( {\sqrt {2x - 1} } \right)}^2}}}{{h\left( {\sqrt {2x + 2h - 1} } \right) + \left( {\sqrt {2x - 1} } \right)}}$
$\Rightarrow f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{2x + 2h - 1 - 2x + 1}}{{h\left( {\sqrt {2x + 2h - 1} } \right) + \left( {\sqrt {2x - 1} } \right)}}$
$\Rightarrow f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{2h}}{{h\left( {\sqrt {2x + 2h - 1} } \right) + \left( {\sqrt {2x - 1} } \right)}}$
$\Rightarrow f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{2}{{\left( {\sqrt {2x + 2h - 1} } \right) + \left( {\sqrt {2x - 1} } \right)}}$
now we can apply the limit, we will have definite value
$\Rightarrow f'(x) = \dfrac{2}{{\left( {\sqrt {2x + 2\left( 0 \right) - 1} } \right) + \left( {\sqrt {2x - 1} } \right)}}$
$\Rightarrow f'(x) = \dfrac{2}{{\left( {\sqrt {2x - 1} } \right) + \left( {\sqrt {2x - 1} } \right)}}$
$\Rightarrow f'(x) = \dfrac{2}{{2\left( {\sqrt {2x - 1} } \right)}}$
$\Rightarrow f'(x) = \dfrac{1}{{\sqrt {2x - 1} }}$ . This is the required answer
Therefore, the correct answer is “ $f'(x) = \dfrac{1}{{\sqrt {2x - 1} }}$ ”.

Note : We can solve this if we know the differentiation formula. That is the differentiation of ${x^n}$ with respect to ‘x’ is $\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{n - 1}}$ .
We have, $f(x) = \sqrt {2x - 1} = {\left( {2x - 1} \right)^{\dfrac{1}{2}}}$ . Differentiation with respect to ‘x’ and using the formula we have,
$\Rightarrow f'(x) = \dfrac{1}{2}{\left( {2x - 1} \right)^{\dfrac{1}{2} - 1}}\dfrac{{d\left( {2x} \right)}}{{dx}}$
$\Rightarrow f'(x) = \dfrac{1}{2}{\left( {2x - 1} \right)^{ - \dfrac{1}{2}}}.2$
$\Rightarrow f'(x) = {\left( {2x - 1} \right)^{ - \dfrac{1}{2}}}$
$\Rightarrow f'(x) = \dfrac{1}{{\sqrt {2x - 1} }}$.
Thus in both the cases we have the same answer. But they particularly asked us to solve using the limit definition so we need to solve the above one.