Finding dx/dt & d^2x/dt^2 for x+e^x=t

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Homework Help Overview

The problem involves finding the first and second derivatives of x with respect to t, given the equation x + e^x = t. The subject area pertains to calculus, specifically implicit differentiation and derivatives.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the differentiation of the equation and the implications of treating x as a function of t. Questions arise regarding the correctness of the second derivative and the application of the chain rule in this context.

Discussion Status

Participants are exploring the differentiation process and questioning the assumptions made in the initial attempts. Some guidance has been offered regarding the need for implicit differentiation and the correct application of the chain rule.

Contextual Notes

There is an emphasis on the relationship between x and t, with participants noting that x is not a constant but a function of t, which affects the differentiation process. The discussion reflects on the need for clarity in applying calculus principles to the problem.

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Homework Statement



x+e^x=t. find dx/dt and d^2x/dt^2

Homework Equations





The Attempt at a Solution


dt/dx = 1 + e^x
Therefore dx/dt = (1+e^x)^-1 This is right.
d^2x/dt^2 = 0. This is wrong. Why?
 
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coverband said:

Homework Statement



x+e^x=t. find dx/dt and d^2x/dt^2

Homework Equations


The Attempt at a Solution


dt/dx = 1 + e^x
Therefore dx/dt = (1+e^x)^-1 This is right.
d^2x/dt^2 = 0. This is wrong. Why?

Because x is a funtion of t, not constant. You need to use the chain rule. And it would also have been more in the spirit of the problem to differentiate the original equation implicitly instead of solving for t.
 
yea, its not 0

how do you differentiate

(1+ex)-1 ??
 
Well, given that x=x(t) (x being a function of t) you get [tex]-(1+e^x)^{-2}*e^x*\frac{dx}{dt}[/tex] And you've already calculated dx/dt.
 

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