Question

How do you find the derivative of $f\left( x \right) = \dfrac{1}{{x - 1}}$?

Answer 1

Here, we will use the limit definition formula for the given functions. Then by substituting the limits, we will find the derivative of the function. Differentiation is a method of finding the derivative of the function and finding the rate of change of a function with respect to one variable.

Formula Used:
Limit definition is given by $f'\left( x \right) = \mathop {\lim }\limits_{\Delta x \to 0} \dfrac{{f\left( {x + \Delta x} \right) - f\left( x \right)}}{{\Delta x}}$.

Complete Step by Step Solution:
We are given a function
$f\left( x \right) = \dfrac{1}{{x - 1}}$ …………………………$\left( 1 \right)$
Now, we will find $f\left( {x + \Delta x} \right)$, so we get
$\Rightarrow f\left( {x + \Delta x} \right) = \dfrac{1}{{x + \Delta x - 1}}$ …….…………………….$\left( 2 \right)$
Now, we will find the derivative of $\dfrac{1}{{x - 1}}$ using the limit definition.
Now, by substituting the equation $\left( 1 \right)$ and equation $\left( 2 \right)$ in the limit definition $f'\left( x \right) = \mathop {\lim }\limits_{\Delta x \to 0} \dfrac{{f\left( {x + \Delta x} \right) - f\left( x \right)}}{{\Delta x}}$, we get
$\Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{\Delta x \to 0} \dfrac{{\dfrac{1}{{x + \Delta x - 1}} - \dfrac{1}{{x - 1}}}}{{\Delta x}}$ .
Now, by taking the L.C.M for the equation in the numerator, we get
$\Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{\Delta x \to 0} \dfrac{{\dfrac{{x - 1}}{{\left( {x + \Delta x - 1} \right)\left( {x - 1} \right)}} - \dfrac{{x + \Delta x - 1}}{{\left( {x + \Delta x - 1} \right)\left( {x - 1} \right)}}}}{{\Delta x}}$ .
$\Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{\Delta x \to 0} \dfrac{{\dfrac{{x - 1 - x - \Delta x + 1}}{{\left( {x + \Delta x - 1} \right)\left( {x - 1} \right)}}}}{{\Delta x}}$ .
Now, by simplifying the equation, we get
$\Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{\Delta x \to 0} \dfrac{{\dfrac{{ - \Delta x}}{{\left( {x + \Delta x - 1} \right)\left( {x - 1} \right)}}}}{{\Delta x}}$
$\Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{\Delta x \to 0} \dfrac{{ - \Delta x}}{{\left( {x + \Delta x - 1} \right)\left( {x - 1} \right)}} \times \dfrac{1}{{\Delta x}}$ .
By dividing the terms, we get
$\Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{\Delta x \to 0} \dfrac{{ - 1}}{{\left( {x + \Delta x - 1} \right)\left( {x - 1} \right)}}$
Now, by substituting the limits, we get
$\Rightarrow f'\left( x \right) = \dfrac{{ - 1}}{{\left( {x + 0 - 1} \right)\left( {x - 1} \right)}}$
Subtracting the terms in the denominator, we get
$\Rightarrow f'\left( x \right) = \dfrac{{ - 1}}{{\left( {x - 1} \right)\left( {x - 1} \right)}}$
$\Rightarrow f'\left( x \right) = \dfrac{{ - 1}}{{{{\left( {x - 1} \right)}^2}}}$

Therefore, the derivative of $\dfrac{1}{{x - 1}}$ using the limit definition is $\dfrac{{ - 1}}{{{{\left( {x - 1} \right)}^2}}}$.

Additional information:
We know that the reverse process of differentiation is called antidifferentiation or integration. We should remember some rules in differentiation which include that the derivative of a constant multiplied by a function is equal to the constant multiplied by the derivative of the function. The derivative of a constant is always zero since zero is a constant; its derivative is zero.

Note:
We can also solve this question using an alternate method. We are given the function in the ratio of two functions, so we can use the Quotient rule to find the derivative of the function.
Now, we will find the derivative of the given function $f\left( x \right) = \dfrac{1}{{x - 1}}$ by using the quotient rule.
Quotient rule is given by the formula $f'\left( {\dfrac{u}{v}} \right) = \dfrac{{vu' - uv'}}{{{v^2}}}$.
$f'\left( x \right) = \dfrac{{\left( {x - 1} \right)\dfrac{d}{{dx}}\left( 1 \right) - 1\dfrac{d}{{dx}}\left( {x - 1} \right)}}{{{{\left( {x - 1} \right)}^2}}}$
Multiplying the terms, we get
$\Rightarrow f'\left( x \right) = \dfrac{{\left( {x - 1} \right)\dfrac{d}{{dx}}\left( 1 \right) - \dfrac{{dx}}{{dx}} - \dfrac{d}{{dx}}\left( 1 \right)}}{{{{\left( {x - 1} \right)}^2}}}$
Now using the differentiation formula $\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$ and $\dfrac{d}{{dx}}\left( k \right) = 0$, we get
$\Rightarrow f'\left( x \right) = \dfrac{{\left( {x - 1} \right)\left( 0 \right) - 1\left( 1 \right)}}{{{{\left( {x - 1} \right)}^2}}}$
Simplifying the equation, we get
$\Rightarrow f'\left( x \right) = \dfrac{{ - 1}}{{{{\left( {x - 1} \right)}^2}}}$
Therefore, the derivative of $\dfrac{1}{{x - 1}}$ is $\dfrac{{ - 1}}{{{{\left( {x - 1} \right)}^2}}}$.