Derivative of a composite function

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Discussion Overview

The discussion revolves around the differentiation of a composite function, specifically focusing on how to handle the term \((y')^2\) when calculating the derivative \(dF(y)/dy\) for the function \(F(y) = y^3 + (y')^2\). The scope includes mathematical reasoning and technical explanation related to calculus and the chain rule.

Discussion Character

  • Technical explanation, Mathematical reasoning

Main Points Raised

  • One participant seeks clarification on how to differentiate the term \((y')^2\) when calculating \(dF(y)/dy\).
  • Another participant reminds them of the chain rule, suggesting they may be struggling with differentiating that term.
  • A different participant explains that the derivative of \((y')^2\) with respect to \(x\) is \(2y' y''\), applying the chain rule to the function \(y\) as a function of \(x\).
  • One participant clarifies their intention to calculate \(dF(y)/dy\) rather than \(dF(y)/dx\), indicating a focus on the derivative with respect to \(y\).
  • Another participant states that differentiating \(d(y)^2/dy\) results in \(2y\), which does not involve \(x\).

Areas of Agreement / Disagreement

The discussion includes multiple viewpoints on how to approach the differentiation, and there is no consensus on the best method to handle the term \((y')^2\) in the context of \(dF(y)/dy\).

Contextual Notes

Participants have not resolved the relationship between derivatives with respect to \(y\) and \(x\), and there are assumptions about the definitions of \(y'\) and \(y''\) that remain unaddressed.

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y =f(x), and y'=df(x)/dx, F(y)=y^3+(y')^2
how to deal with the (y')^2 when i calculate dF(y)/dy?
thanks.
 
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Do you remember the chain rule? Are you having trouble differentiating that term?
 
If u(x) is a fuction of x, then, by the chain rule, the derivative of [itex]u^2[/itex] is 2u times the derivative of u.

If y is a function of x, then the derivative of [itex](y')^2[/itex], with respect to x, is [itex]2y'[/itex] times the derivative of y' which is, of course, y''. That is, the derivative if [itex](y')^2[/itex] is [itex]2y' y''[/itex].
 
HallsofIvy, i want to calculate dF(y)/dy, not dF(y)/dx, but you remand me sth. thank you.
 
Well, that doesn't require any mention of x at all then! The derivative of [itex]d(y)^2/dy= 2y[/itex].
 
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