How do I find the derivative of y = x²log3x?

Question

it's not log"3x", it's log 3 of x.

Answer 1

You can transform log3 x into ln x / ln 3, so then y becomes

y = x^2 * ln x / ln 3

Using the multiplication rule (d(ab)/dx = da/dx * b + a * db/dx), we get

dy/dx = ( d(x^2)/dx * ln x + x ^2 * d(ln x)/dx) / ln 3
= (2x ln x + 1/x * x^2) / ln 3
dy/dx = (2x ln x + x) / ln 3


Addendum: See http://ltcconline.net/greenl/courses/116/ExpLog/logDerivative.htm for more on taking derivatives of logarithms.

Answer 2

In the following, log is taken to mean log to base 3:-
y = ( x ² ) ( log x )
dy/dx = 2x log x + (1/x) x²
dy/dx = 2x log x + x