Question

Find derivative of $f(x) = \dfrac{1}{x}$using the first principle of differentiation.

Answer 1

The given question requires us to find the derivative of a function using the first principle of differentiation. The first principle of differentiation helps us evaluate the derivative of a function using limits. Calculating the derivative of a function using the first principle of differentiation may be a tedious task. We may employ identities and tricks to calculate the limits and evaluate the required derivative.

Complete step by step answer:
We have to evaluate the derivative of $f(x) = \dfrac{1}{x}$ using the first principle of differentiation.

According to the first principle of differentiation, the derivative of a function can be evaluated by calculating the limit $f'\left( x \right){\text{ = }}\mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f{\text{(x)}}}}{h}$ .

So, the derivative of the function $f(x) = \dfrac{1}{x}$ can be calculated by the first rule of differentiation as:
$f'(x){\text{ = }}\mathop {\lim }\limits_{h \to 0} \left[ {\dfrac{{\dfrac{1}{{x + h}} - \dfrac{1}{x}}}{h}} \right]$

Taking the LCM of the fractions, we get,
$f'(x){\text{ = }}\mathop {\lim }\limits_{h \to 0} \left[ {\dfrac{{x - \left( {x + h} \right)}}{{h(x)(x + h)}}} \right]$

Opening the brackets and simplifying further, we get,
$f'(x){\text{ = }}\mathop {\lim }\limits_{h \to 0} \left[ {\dfrac{{x - x - h}}{{h(x)(x + h)}}} \right]$

Cancelling the like terms with opposite signs,
$f'(x){\text{ = }}\mathop {\lim }\limits_{h \to 0} \left[ {\dfrac{{ - h}}{{h(x)(x + h)}}} \right]$

Cancelling the numerator and denominator and simplifying the limit further,
$f'(x){\text{ = }}\mathop {\lim }\limits_{h \to 0} \left[ {\dfrac{{ - 1}}{{(x)(x + h)}}} \right]$

Now, On applying the limit, we get
$f'(x){\text{ = }}\left[ {\dfrac{{ - 1}}{{{x^2}}}} \right]$

So, the derivative of the function $f(x) = \dfrac{1}{x}$ is $f'(x){\text{ = }}\left( {\dfrac{{ - 1}}{{{x^2}}}} \right)$ .

Note: The derivative of the given function can also be calculated by using the power rule of differentiation. According to the power rule of differentiation, the derivative of ${x^n}$ is $n{x^{n - 1}}$ .
So, going by the power rule of differentiation, the derivative of $f(x) = \dfrac{1}{x} = {x^{ - 1}}$ is $( - 1){x^{ - 2}}$. So, the derivative of the given function $f(x) = \dfrac{1}{x}$ is $\dfrac{1}{{{x^2}}}$.