What is the derivative of #log_3x#?

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Answer 1 · 2018-03-31T20:40:01

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

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).#

Answer 2 · 2018-03-31T21:14:02

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

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)#