Question

How do you differentiate ${\log _2}\left( x \right)$?

Answer 1

Hint: Here we just need to see that the base of the logarithm function is not $e$ and it is $2$.
Whenever we are given such logarithmic function we just need to apply the formula which we represent by $\dfrac{d}{{dx}}\left( {{{\log }_a}x} \right) = \dfrac{1}{{x.\ln \left( a \right)}}$

Complete step by step answer:
Here we are given to differentiate the term which is given as ${\log _2}\left( x \right)$ and here the base of the logarithmic function is not $e$ and it is $2$
We need to know that when we have such type of problems we need to apply the formula that states:
$\dfrac{d}{{dx}}\left( {{{\log }_a}x} \right) = \dfrac{1}{{x.\ln \left( a \right)}}$
It is actually the same formula as we have just changed the base and we know that ${\log _a}x = \dfrac{{\ln \left( x \right)}}{{\ln \left( a \right)}}$
So we have simply substituted this value in the above formula and got the differentiation as we have written above.
So let us apply this formula over ${\log _2}\left( x \right)$ we will get:
$\dfrac{d}{{dx}}\left( {{{\log }_a}x} \right) = \dfrac{1}{{x.\ln \left( a \right)}}$
$\dfrac{d}{{dx}}\left( {{{\log }_2}x} \right)$
Now if we equate this expression to which we have been given we can compare and say that $a = 2$ and we will get:
\[\dfrac{d}{{dx}}\left( {{{\log }_2}x} \right) = \dfrac{1}{{x.\ln \left( 2 \right)}}\]

Hence we can say that the derivative of ${\log _2}\left( x \right)$ is \[\dfrac{1}{{x.\ln \left( 2 \right)}}\].

Note: Here the student must know all the properties of logarithm like:
$
  \log \left( {ab} \right) = \log a + \log b \\
  \log \left( {\dfrac{a}{b}} \right) = \log a - \log b \\
 $
$
  {\log _a}b = \dfrac{{\log b}}{{\log a}} \\
  \log {m^n} = n\log m \\
 $