What is the derivative of \[{\log _3}x\]?
Answer
521.1k+ views
Hint: Derivative of a function gives the rate of change of the function value with respect to change in its argument value. To find the derivative of \[{\log _3}x\], we will first apply the base change rule of logarithm and change the base of \[{\log _3}x\] from \[3{\text{ to e}}\]. Then we will apply the formula for the derivative of \[\ln x\] and find the derivative of \[{\log _3}x\].
Complete step by step answer:
Let,
\[y = {\log _3}x\]
Now we know from the base change property of logarithm that,
\[{\log _3}x = \dfrac{{{{\log }_e}x}}{{{{\log }_e}3}}\]
So, using this we get;
\[ \Rightarrow y = \dfrac{{{{\log }_e}x}}{{{{\log }_e}3}}\]
Now differentiating both sides we get;
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\dfrac{{{{\log }_e}x}}{{{{\log }_e}3}}} \right)\]
Now we know \[{\log _e}3\] is a constant. So, we will take it out of the differentiation sign.
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{{{\log }_e}3}} \times \dfrac{d}{{dx}}\left( {{{\log }_e}x} \right)\]
Now we know that, \[\dfrac{{d\ln x}}{{dx}} = \dfrac{1}{x}\], so we get;
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{x{{\log }_e}3}}\]
We can also write it as;
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{x\ln 3}}\]
Note:
One mistake that most of the students commit in these types of questions is that they simply apply the formula that \[\dfrac{d}{{dx}}\log x = \dfrac{1}{x}\]. This is because this formula is valid only when the base is \[e\]. But in the question the base of logarithm is \[3\]. So, we have to change the base first and then do the differentiation.
Complete step by step answer:
Let,
\[y = {\log _3}x\]
Now we know from the base change property of logarithm that,
\[{\log _3}x = \dfrac{{{{\log }_e}x}}{{{{\log }_e}3}}\]
So, using this we get;
\[ \Rightarrow y = \dfrac{{{{\log }_e}x}}{{{{\log }_e}3}}\]
Now differentiating both sides we get;
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\dfrac{{{{\log }_e}x}}{{{{\log }_e}3}}} \right)\]
Now we know \[{\log _e}3\] is a constant. So, we will take it out of the differentiation sign.
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{{{\log }_e}3}} \times \dfrac{d}{{dx}}\left( {{{\log }_e}x} \right)\]
Now we know that, \[\dfrac{{d\ln x}}{{dx}} = \dfrac{1}{x}\], so we get;
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{x{{\log }_e}3}}\]
We can also write it as;
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{x\ln 3}}\]
Note:
One mistake that most of the students commit in these types of questions is that they simply apply the formula that \[\dfrac{d}{{dx}}\log x = \dfrac{1}{x}\]. This is because this formula is valid only when the base is \[e\]. But in the question the base of logarithm is \[3\]. So, we have to change the base first and then do the differentiation.
Recently Updated Pages
Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Class 12 Question and Answer - Your Ultimate Solutions Guide

Master Class 9 General Knowledge: Engaging Questions & Answers for Success

Master Class 9 Maths: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Sulphuric acid is known as the king of acids State class 12 chemistry CBSE

Why is the cell called the structural and functional class 12 biology CBSE

A dentist uses a small mirror that gives a magnification class 12 physics CBSE

