Finding the derivative, quotient rule with natural log function.

Click For Summary
SUMMARY

The discussion focuses on finding the derivative of the function y = 1 - (3ln(7x))/x^4 using the quotient rule. The user initially applied the quotient rule incorrectly, leading to a miscalculation in the simplification process. The error identified was the omission of a minus sign, which resulted in an incorrect expression for the derivative. The correct derivative is y' = -3x^3 - 4x^3, simplifying to y' = -7x^3.

PREREQUISITES
  • Understanding of calculus, specifically differentiation techniques.
  • Familiarity with the quotient rule for derivatives.
  • Knowledge of logarithmic differentiation, particularly with natural logarithms.
  • Ability to simplify algebraic expressions involving derivatives.
NEXT STEPS
  • Review the quotient rule for derivatives in calculus.
  • Practice problems involving logarithmic differentiation with natural logs.
  • Learn how to identify and correct common errors in derivative calculations.
  • Explore advanced derivative techniques, such as implicit differentiation.
USEFUL FOR

Students studying calculus, particularly those learning about derivatives and the quotient rule, as well as educators looking for examples of common mistakes in derivative calculations.

HHenderson90
Messages
9
Reaction score
0

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?
 
Physics news on Phys.org
HHenderson90 said:

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.
 

Similar threads

  • · Replies 25 ·
Replies
25
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 24 ·
Replies
24
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
5
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K