How do you find the derivative of #y=log_4 2x#?

Question

1574 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-25T19:02:51

#y'=1/(xln4)#

Rewrite #y# using the change of base formula.

#y=ln(2x)/ln4#

#y'=1/ln4*d/dx(ln(2x))#

Recall that #1/ln4# is just a constant.

To find #d/dx=(ln(2x))#, use the chain rule.

#ln(u)=1/u*(du)/dx#

Thus,

#d/dx(ln(2x))=1/(2x)d/dx(2x)=2/(2x)=1/x#

Plug this back into the #y'# expression.

#y'=1/(xln4)#