Question

The differentiation of \[\dfrac{d}{{dx}}\left( {\dfrac{1}{x}} \right)\] when $x \ne 0$is
A) $ - \dfrac{1}{{{x^2}}}$
B) $\dfrac{1}{{{x^2}}}$
C) Not Possible
D) 1

Answer 1

Hint: In this question, we have to differentiate the given function with respect to $x$. We can solve this question by using two methods- one is the Quotient rule and the other is the power rule. In the Quotient rule, we use the direct formula of differentiation and differentiate it.
Power rule- In the power rule firstly, we take “x” in the numerator by taking the power of “x” as negative and then use the power rule of the differential equation to solve the question.

Formula used:
Quotient rule:
$\dfrac{d}{{dx}}(\dfrac{{f(x)}}{{g(x)}}) = \dfrac{{g(x)\dfrac{{df(x)}}{{dx}} - f(x)\dfrac{{dg(x)}}{{dx}}}}{{{{(g(x))}^2}}}$
Here $f(x)$ and $g(x)$ is the function of “$x$”.
Power rule:
$\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$
Here $n$ is the integer.

Complete step-by-step solution:
Let us first calculate the differential equation by using the quotient rule.
Let $y = \dfrac{d}{{dx}}\left( {\dfrac{1}{x}} \right) - - - (1)$
Here we have,
$
  f(x) = 1 \\
  g(x) = x \\
 $
Now differentiate the equation $(1)$ we get;
$y = \dfrac{{x\dfrac{d}{{dx}}(1) - (1)\dfrac{d}{{dx}}(x)}}{{{x^2}}}$
Here as we know, the differentiation of the constant term “$1$” is zero and the differentiation of “$x$” is $1$. Therefore, we have;
$
  y = \dfrac{{0 - 1}}{{{x^2}}} \\
   \Rightarrow y = - \dfrac{1}{{{x^2}}} \\
 $
Now, let us solve this question by using the power rule;
$y = \dfrac{d}{{dx}}\left( {\dfrac{1}{x}} \right)$
We can write the above equation as;
$y = \dfrac{d}{{dx}}\left( {{x^{ - 1}}} \right)$
Using Power rule and differentiate it, we get;
$
  y = - 1({x^{ - 1 - 1}}) \\
   \Rightarrow y = - {x^{ - 2}} \\
   \Rightarrow y = - \dfrac{1}{{{x^2}}} \\
 $

Hence, option A) is the correct answer.

Note: In the differentiation, many students get confused that the differentiation of constant function is equal to $1$ which makes their answer incorrect because always remember that the differentiation of constant function is equal to $0$ and differentiation of function “$x$” is equal to $1$.
Do not confuse the product and the quotient rules. In the product rule, we use the addition sign and, in the quotient rule, we use the subtraction sign to differentiate the question.