How do I find v' for functions v=x+y using implicit differentiation?

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Discussion Overview

The discussion revolves around finding the derivative v' for the function v = x + y using implicit differentiation. Participants explore the application of differentiation techniques, including the chain rule and product rule, in the context of functions dependent on x.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant asks how to find v' for the function v = x + y.
  • Another participant suggests differentiating the definition of v with respect to x.
  • A participant clarifies that v' refers to the derivative of v with respect to x and applies the chain rule, resulting in dv/dx = 1 + dy/dx, noting that dy/dx depends on the function y in relation to x.
  • A later reply questions whether dy/dx is the only term involved, recalling a different approach from a textbook that includes (dy/dx)(dv/dx).
  • Another participant introduces a new question regarding the differentiation of the expression d(e^u dy/dx)/du, mentioning confusion about the appearance of dx/du in the product rule application.
  • A participant responds by explaining that the term dx/du arises from the chain rule.

Areas of Agreement / Disagreement

Participants express varying levels of understanding regarding the application of differentiation rules, and there is no consensus on the interpretation of certain derivative expressions. The discussion remains unresolved on some aspects of the differentiation process.

Contextual Notes

Some assumptions about the function y and its relationship to x are not explicitly stated, and the discussion does not resolve the specific conditions under which dy/dx is defined.

daster
Say we have two functions of x, v and y, such that v=x+y. How can I find v'?
 
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Why doncha differentiate its definition wrt to "x"??

Daniel.

PS.Tell me what u get.
 
Since v and y are functions of x, I presume by " v' " you mean "the derivative of v with respect to x", i.e. dv/dx.

Use the chain rule: v= x+ y so dv/dx= dx/dx+ dy/dx= 1+ dy/dx. What dy/dx is depends, of course, on what function y is of x.
 
Oh, so it's only dy/dx? Cause I remember my book doing something like (dy/dx)(dv/dx) or something. Thanks HallsofIvy. :smile:

Another question. How do I find d(e^u dy/dx)/du, where u and y are functions of x?
My book says:

[tex]e^u \frac{dy}{dx} + e^u \frac{d^2y}{dx^2} \cdot \frac{dx}{du}[/tex]

I understand this is the product rule, but where'd the dx/du come from?
 
Last edited by a moderator:
It comes from the chain rule again:

[tex]\frac{d}{du} \frac{dy}{dx}=\frac{d^2y}{dx^2}\frac{dx}{du}[/tex]
 
Dank je. :smile:
 
Graag gedaan. :biggrin:
 

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