Question

How do you find the derivative of $\dfrac{1}{{2x}}$ ?

Answer 1

Hint : Use the reciprocal rule of derivation i.e. ${\left[ {\dfrac{1}{{u(x)}}} \right]^\prime } = \dfrac{{u'(x)}}{{u{{(x)}^2}}}$ to solve the above problem .
Formula:
${\left[ {\dfrac{1}{{u(x)}}} \right]^\prime } = \dfrac{{u'(x)}}{{u{{(x)}^2}}}$
$\dfrac{d}{{dx}}\left[ x \right] = 1$

Complete step-by-step answer :
Given a function $\dfrac{1}{{2x}}$ let it be $f(x)$
$f(x) = \dfrac{1}{{2x}}$
We have to find the first derivative of the above equation
$\dfrac{d}{{dx}}\left[ {f(x)} \right] = f'(x) \\\ f'(x) = \dfrac{d}{{dx}}\left[ {\dfrac{1}{{2x}}} \right] \\\$
Differentiation is linear So we can differentiate summands easily and pull out the constant factors
$f'(x) = \dfrac{1}{2}\dfrac{d}{{dx}}\left[ {\dfrac{1}{x}} \right]$
Now applying the reciprocal rule ${\left[ {\dfrac{1}{{u(x)}}} \right]^\prime } = \dfrac{{u'(x)}}{{u{{(x)}^2}}}$
$= - \dfrac{{\dfrac{{\dfrac{d}{{dx}}\left[ x \right]}}{{{x^2}}}}}{2}$
The derivative of differentiation variable is 1
$= - \dfrac{1}{{2{x^2}}}$
So, the correct answer is “- $\dfrac{1}{{2{x^2}}}$ ”.

Note : 1. Calculus consists of two important concepts one is differentiation and other is integration.
2.What is Differentiation?
It is a method by which we can find the derivative of the function .It is a process through which we can find the instantaneous rate of change in a function based on one of its variables.
Let y = f(x) be a function of x. So the rate of change of $y$ per unit change in $x$ is given by:
$\dfrac{{dy}}{{dx}}$ .