Understanding Implicit Differentiation: Solving for Second Order Derivatives

Click For Summary

Homework Help Overview

The discussion revolves around implicit differentiation, specifically focusing on finding the second order derivative of the equation 3y² + 8y = 3x. Participants are exploring the relationship between the variables and the implications of expressing derivatives in terms of x and y.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to express derivatives in terms of both x and y, with some questioning the validity of expressing y solely in terms of x. There are attempts to compute the second derivative, with varying levels of success and understanding.

Discussion Status

The discussion is active, with participants providing insights and corrections regarding the computation of derivatives. There is a recognition of the need to substitute expressions derived from the original equation into the second derivative calculation, and some participants are beginning to grasp the connections being made.

Contextual Notes

There is an ongoing debate about the appropriateness of expressing y as a function of x, given the implicit nature of the original equation. Some participants note that this could complicate the differentiation process.

Duncan1382
Messages
9
Reaction score
0
Find [tex]d^2/dx^2(3y^2+8y=3x)[/tex]

I managed to get [tex]dy/dx = 3 / (6y + 8)[/tex] but I have no clue where to go from here.

According to WolfRamAlpha, the answer is [tex]-27/(4(16 + 9x)(4 + 3y))[/tex], but since dy/dx doesn't have any x value in it, I don't see how the derivative of it would.

I've played around with it for a long time, and I just can't get it. Help please?
 
Physics news on Phys.org
Do you need to express in terms of x and y? I don't see what's wrong if you express it in y which is what i would do
 
semc said:
Do you need to express in terms of x and y? I don't see what's wrong if you express it in y which is what i would do

You can't express y in terms of x. It's not a function. You can express it in terms of x and y, but that just makes the math longer.
 
Duncan - WolfRamAlpha mad a substitution in the denominator using the original expression.

Firstly - did you compute the second derivative correctly? You should have yielded:

[tex] y' =\frac{3}{8+6 \cdot y}[/tex]
and
[tex] y'' = -\frac{6 \cdot (y')^2}{8+6 \cdot y}[/tex]

Now, using the first into the second

[tex] y'' = -\left (\frac{3}{8+6 \cdot y} \right)^2 \left (\frac{6}{8+6 \cdot y} \right ) = \frac{-54}{8 \cdot (4+3 \cdot y)^3}[/tex]

Now, using the fact that [itex]3y^2+8y=3x[/itex], you need to show yourself that [itex](4+3y)^2 = 9x +16[/itex]. Use this in above to yield result.
 
Oh. Now I get it.

Thank you.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 25 ·
Replies
25
Views
2K
  • · Replies 24 ·
Replies
24
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 24 ·
Replies
24
Views
3K