Implicit Differentiation: Finding d^2y/dx^2 at a Given Point

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Homework Help Overview

The problem involves finding the second derivative, d²y/dx², of the implicit function defined by the equation y³ + y = 2 cos x at the point (0,1). The subject area is implicit differentiation within calculus.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to differentiate the equation to find dy/dx and expresses uncertainty about how to proceed to find d²y/dx² and the relevance of the point (0,1). Some participants suggest differentiating both sides of the equation again, using the chain rule, and provide guidance on focusing on the equation involving y' for the second differentiation.

Discussion Status

The discussion is ongoing, with participants providing guidance on the differentiation process. There is a recognition of the correct first derivative, and suggestions are made for the next steps without reaching a consensus on the final approach.

Contextual Notes

There is a mention of using the point (0,1) in the context of finding the second derivative, but the implications of this point are not fully explored. The participants are navigating the complexities of implicit differentiation and the application of the chain rule.

athamz
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Homework Statement


find the d^2y/dx^2 if y^3 + y = 2 cos x at the point (0,1)


Homework Equations





The Attempt at a Solution



my dy/dx = (-2 sin x)/(3y^2 + 1)

I don't know how to find d^2y/dx^2?
And when and how will I use the oint (0,1)?
 
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Differentiate both sides of the equation twice with respect to x, using chain rule on the left-hand size (differentiate with respect to y first than with respect to x).

ehild
 
oh.. so I have..

3y^2 * y' + y' = -2 sin x

y'(3y^2 + 1) = -2 sin x

y' = (-2 sin x)/(3y^2 + 1)

Right?
 
It is right. Get the second differential. But differentiate both sides of the equation y'(3y^2 + 1) = -2 sin x instead of the last one. It will be an easier process.

ehild
 

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