Question

What is the differentiation of $\log 2x$ ?

Answer 1

Hint : In the given problem, we are required to differentiate $\log 2x$ with respect to x. Since, $\log 2x$ is a composite function, we will have to apply the chain rule of differentiation in the process of differentiating $\log 2x$ . So, differentiation of $\log 2x$ with respect to x will be done layer by layer using the chain rule of differentiation. Also derivatives of 2x with respect to x must be remembered in order to solve the given problem.

Complete step by step solution:
So, we have, $\dfrac{d}{{dx}}\left( {\log 2x} \right)$
Keeping the expression inside the logarithmic function inside the bracket, we get,
$=$ $\dfrac{d}{{dx}}\left( {\log \left( {2x} \right)} \right)$
Now, Let us assume $u = 2x$. So substituting $2x$ as $u$, we get,
$=$ $\dfrac{d}{{dx}}\left( {\log u} \right)$
Now, we know that differentiation of logarithmic function $\log x$ with respect to x is $\left( {\dfrac{1}{x}} \right)$. So, we get,
$=$ $\dfrac{1}{u} \times \dfrac{{du}}{{dx}}$
Now, putting back $u$as $2x$, we get,
$=$ $\dfrac{1}{{2x}} \times \dfrac{{d\left( {2x} \right)}}{{dx}}$ because $\dfrac{{du}}{{dx}} = \dfrac{{d\left( {2x} \right)}}{{dx}}$
Now, we take the constants out of the differentiation. So, we get,
$=$ $\dfrac{2}{{2x}} \times \dfrac{{d\left( x \right)}}{{dx}}$
Now, we know that the derivative of x with respect to x is $1$. Hence, we get,
$=$ $\dfrac{2}{{2x}} \times 1$
Cancelling the common factors in numerator and denominator, we get,
$=$ $\dfrac{1}{x}$
So, the derivative of $\log 2x$ with respect to x is $\dfrac{1}{x}$.
So, the correct answer is “$\dfrac{1}{x}$”.

Note : The chain rule of differentiation involves differentiating a composite by introducing new unknowns to ease the process and examine the behaviour of function layer by layer. The derivative of the basic logarithmic function $\log x$ with respect to x is $\dfrac{1}{x}$. The power rule of differentiation is as follows: $\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{n - 1}}$.