Question about using derivatives

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Homework Help Overview

The problem involves finding the approximate percentage changes in the function y = f(x) = x² resulting from a 2% increase in x. Participants are discussing the use of derivatives in this context.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply the derivative to find the percentage change but questions the presence of the variable x in their result. Some participants suggest that a second derivative might be relevant, while others clarify that only the first derivative is necessary.

Discussion Status

Participants are exploring different interpretations of the problem, particularly regarding the use of derivatives and the role of the variable x in the solution. There is no explicit consensus, but some guidance has been offered regarding the correct derivative to use.

Contextual Notes

There seems to be confusion regarding the parameter 't' mentioned in the original poster's attempt, as well as the requirement to express the answer without variables. This may be influencing the discussion.

Cbray
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Homework Statement


Find the approximate percentage changes in the given function y = f(x) that will result from an increase of 2%
y = x2

Homework Equations





The Attempt at a Solution


dy/dt = dx/dt * d/dx * x2
dy/dt = dx/dt * 2x
dy/dt = 2/100 * 2x
dy/dt = 4x% ? I don't know if I did this right but apparently in the back of the book I'm not suppose to have the x in there, why not?
 
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Would it be because your supposed to take the second derivative of it? That would make it just 2. I think as the question wants no variables in it too.
 
hvidales said:
Would it be because your supposed to take the second derivative of it? That would make it just 2. I think as the question wants no variables in it too.

I don't think that is correct...
 
What is 't'? You have [itex]y = x^2[/itex] and x seems to have no dependence on a parameter 't'. You simply need to determine [itex]{{dy}\over{dx}}[/itex]

And no, there are no second derivatives involved.
 
Pengwuino said:
What is 't'? You have [itex]y = x^2[/itex] and x seems to have no dependence on a parameter 't'. You simply need to determine [itex]{{dy}\over{dx}}[/itex]

And no, there are no second derivatives involved.

Do you mind filling me in a bit more, i.e. is my working wrong apart from that
 
Last edited:
Cbray said:
Do you mind filling me in a bit more, i.e. is my working wrong apart from that

There is not really a working apart from that. So no, it's not correct.
 
[solved]
dy/dx=2x
dy/y=2x dx/x2
dy/y=2 * 2/100
dy/y = 4%
 

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