• Support PF! Buy your school textbooks, materials and every day products Here!

Finding the derivative.

  • Thread starter chaosblack
  • Start date
  • #1
16
0
1. Homework Statement

Find [tex]\frac{dy}{dx}[/tex] if y = [tex]\sqrt{5u^2 -3}[/tex] and u = [tex]\frac{2x}{3x+1}[/tex]


2. Homework Equations

Chain Rule
[tex]\frac{dy}{dx}[/tex] = [tex]\frac{dy}{du}[/tex] x [tex]\frac{du}{dx}[/tex]


3. The Attempt at a Solution

[tex]\frac{dy}{du}[/tex] = 5u(5u^2 - 3)^-1/2

[tex]\frac{du}{dx}[/tex] = -6x(3x+1)^-2

[tex]\frac{dy}{dx}[/tex] = 5u(5u^2 - 3)^-1/2 x -6x(3x+1)^-2
= -30xu(5u^2 - 3)^-1/2 x (3x+1)^-2
= [tex]\frac{-60x^2}{3x+1}[/tex](5([tex]\frac{2x}{3x+1}[/tex])^2 - 3)^-1/2 x (3x+1)^-2

Is that the final simplified answer?
 

Answers and Replies

  • #2
G01
Homework Helper
Gold Member
2,649
16
Check your work for du/dx. I think you may have an error. You should have a 2 in the numerator, not a -6x.
 
  • #3
16
0
u = [tex]\frac{2x}{3x+1}[/tex]
u = (2x)(3x+1)^-1
u' = -2x(3x+1)^-2 x (3)
u' = -6x(3x+1)^-2

Did I do something wrong?
 
  • #4
CompuChip
Science Advisor
Homework Helper
4,284
47
[tex]\frac{d}{dx}(f(x) g(x)) = \frac{df(x)}{dx} g(x) + f(x) \frac{dg(x)}{dx}[/tex]
You have a term missing.
 
  • #5
16
0
u = (2x)(3x+1)^-1
u' = 2(3x+1)^-1 + (2x)(-1)(3x+1)^2(3)
= 2(3x+1)^-1 - 6x(3x+1)^-2

This?
 
  • #6
G01
Homework Helper
Gold Member
2,649
16
Yes, that's the correct du/dx.
 
  • #7
16
0
dy/dx = 5u(5u^2 - 3)^-1/2 x (2(3x+1)^-1 - 6x(3x+1)^-2)

From here do I just sub in for u?
 
  • #8
G01
Homework Helper
Gold Member
2,649
16
Yep. You got it.
 

Related Threads for: Finding the derivative.

  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
1
Views
820
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
12
Views
1K
  • Last Post
Replies
3
Views
8K
  • Last Post
Replies
9
Views
1K
  • Last Post
Replies
4
Views
831
  • Last Post
Replies
19
Views
2K
Top