Finding the derivative, quotient rule with natural log function.

[HHenderson90]

Homework Statement

Find y' of
y= 1-3ln(7x)/x^4

Homework Equations

The Attempt at a Solution

I used the quotient rule and got:
y'=x^4*d/dx(1-3ln(7x)-(1-3ln(7x)*d/dx(x^4)/(x^4)2

which is: x^4*(0-3*1/7x*7)-(1-3ln(7x))*4x^3/x^8
simplified to: 3x^4/x-1+3ln(7x)*4x^3
3x^3-4x^3+12x^3ln(7x)/x^8
take out the x^3 from the denominator and the numerator and I get the answer:

y=-1+12ln(7x)/x^5
The website I use to do my calculus homework says this is wrong, where is my error?

 

[Dick]

Homework Statement

Find y' of
y= 1-3ln(7x)/x^4

Homework Equations

The Attempt at a Solution

I used the quotient rule and got:
y'=x^4*d/dx(1-3ln(7x)-(1-3ln(7x)*d/dx(x^4)/(x^4)2

which is: x^4*(0-3*1/7x*7)-(1-3ln(7x))*4x^3/x^8
simplified to: 3x^4/x-1+3ln(7x)*4x^3
3x^3-4x^3+12x^3ln(7x)/x^8
take out the x^3 from the denominator and the numerator and I get the answer:

y=-1+12ln(7x)/x^5
The website I use to do my calculus homework says this is wrong, where is my error?

You dropped a minus sign. You should get -3x^3-4x^3. Not 3x^3-4x^3.