Question #829b0

2 Answers
Feb 1, 2018

Given #ln(5x-3y^2)#, #d/dx[ln(5x-3y^2)]=(5-6y)/(5x-3y^2)#

Explanation:

#ln(5x-3y^2)#
First, you derive the inside of the function (#5x-3y^2#)
1. #5-6y#
Second, you put this answer over the inside of the function.
2. #d/dx[ln(5x-3y^2)]=(5-6y)/(5x-3y^2)#.

Feb 1, 2018

Please see below.

Explanation:

.

Let #u=5x-3y^2#

#du=5dx-6ydy#

#d(ln(5x-3y^2))=d(lnu)=1/udu=(5dx-6ydy)/(5x-3y^2)#

If we only differentiate with respect to #x# we will have:

#(du)/dx=5#

#d/dx(ln(5x-3y^2))=d/dx(lnu)= 1/udu=5/(5x-3y^2)#