What is the derivative of #log_3x#?

2 Answers
Mar 31, 2018

#d/dx(log_3x)=1/(xln3)#

Explanation:

We can rewrite #log_3x# as #ln(x)/ln(3)#.

So, we really want #d/dx(lnx)/ln3#. Knowing that #d/dxlnx=1/x#, we get

#d/dx(lnx)/ln3=1/(xln3)#.

This gives rise to the general differentiation rule

#d/dxlog_ax=1/(xlna).#

Mar 31, 2018

#d/dx[log_3(x)]=1/(ln(3)*x)#

Explanation:

There is a rule here:

#d/dx[log_a(x)]=1/((ln(a)*x))*d/dx[x]#

Therefore:

#d/dx[log_3(x)]=1/(ln(3)*x)*d/dx[x]#

The derivative of #x# is just #1#. Therefore:

#d/dx[log_3(x)]=1/(ln(3)*x)#