How to differentiate y = 8 ln x - 9 x with respect to x^2?

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Discussion Overview

The discussion revolves around the differentiation of the function y = 8 ln x - 9 x with respect to x². Participants explore various methods and considerations related to this differentiation, including the implications of variable substitution and the domain of the function.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests introducing a new variable z = x², leading to the transformation of the function into y(z) = 8 ln √z - 9√z for differentiation.
  • Another participant challenges the use of the square root, arguing that it introduces ambiguity regarding the sign of the square root and may complicate finding the inverse function in more complex cases.
  • A method for differentiating f(x) with respect to g(x) is presented, involving the chain rule and the derivatives of f and g, leading to a specific expression for the derivative.
  • One participant asserts that x is always positive in this context due to the presence of ln x, suggesting that there is no ambiguity regarding the square root.
  • Another participant acknowledges the existence of negative logarithms but agrees that assuming x is positive is appropriate for this discussion.
  • A later reply emphasizes that the domain of the function is indeed a subset of (0, ∞), reinforcing the assumption of positivity for x.

Areas of Agreement / Disagreement

Participants generally agree on the positivity of x in the context of the logarithmic function, but there is disagreement regarding the best method for differentiation and the implications of using the square root transformation.

Contextual Notes

There are unresolved considerations regarding the choice of variable substitution and the implications of domain restrictions on the differentiation process.

DavidLiew
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How to differentiate y = 8 ln x - 9 x with respect to x^2?
 
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Introduce the new variable z=x². Then x=√z and y(z)=8ln√z - 9√z. Now differentiate wrt z.
 


Hello, quasar. I don't think using square root function is a good method, since you don't know whether to take the positive or negative square root. For more difficult case, you may not be able to find the inverse function.

To differentiate f(x) w.r.t g(x), just do the following:

[tex]\frac{d f(x)}{d g(x)}[/tex]

[tex]=\frac{d f(x)}{d x}\frac{d x}{d g(x)}[/tex]

[tex]=\frac{d f(x)}{d x}\left(\frac{d g(x)}{d x}\right)^{-1}[/tex]

David, for your problem, you just put:

[tex]f(x)=8 \ln x -9x[/tex] and [tex]g(x) = x^2[/tex], and you will have the answer:

[tex]\frac{\frac{8}{x}-9}{2x}[/tex]

[tex]=\frac{8-9 x}{2 x^2}[/tex]
 


ross_tang, x is always positive in this problem because of the "ln x" appearing in the defining formula for y. So I don't think there is an ambiguity of any sort.
 


Prrrrrety sure the domain of the function the OP is interested in is a subset of ]0,infty[. :)

But thanks for the reminder.
 

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