Question

Question #829b0

Answer 1

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)`.

Answer 2

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)`