Derivative and Limit of an Exponential Function

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Discussion Overview

The discussion revolves around finding the derivative and limit of the function \(f(x)=\left ( e^{x}+x \right )^{^{\frac{1}{x}}}\). Participants explore methods for differentiation and limit evaluation, including the use of natural logarithms and L'Hôpital's Rule.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests using the natural logarithm to simplify the differentiation process of the function.
  • Another participant agrees and outlines the steps to take the natural log of both sides and differentiate implicitly.
  • Some participants propose that the limit as \(x\) approaches infinity of the function is \(e\), based on their calculations that lead to a right-hand side of 1.
  • There is a question about the meaning of implicit differentiation, with an explanation provided that it relates to applying the chain rule in cases where \(y\) is not explicitly defined as a function of \(x\).

Areas of Agreement / Disagreement

Participants generally agree on the approach to finding the derivative and the limit, with some expressing confidence that the limit is \(e\). However, the discussion does not reach a consensus on the final answer or the implications of implicit differentiation.

Contextual Notes

The discussion includes references to indeterminate forms and the application of L'Hôpital's Rule, but does not resolve the mathematical steps or assumptions involved in these processes.

Yankel
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Hello all,

I have a complicated function:

\[f(x)=\left ( e^{x}+x \right )^{^{\frac{1}{x}}}\]

I need to find it's derivative and it's limit when x goes to infinity.

As for the derivative, I thought maybe to use LN, so that I can get rid of the exponent, am I correct?

How should I approach the limit ?

Thank you !
 
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Yes, I would agree that to find the derivative, we can write:

$$y=\left(e^x+x\right)^{\frac{1}{x}}$$

Then take the natural log of both sides:

$$\ln(y)=\frac{1}{x}\ln\left(e^x+x\right)$$

Now implicitly differentiate.

For the limit, logs can come to our aid as well...begin by writing:

$$L=\lim_{x\to\infty}\left(\left(e^x+x\right)^{\frac{1}{x}}\right)$$

Upon taking the natural log of both sides, we obtain:

$$\ln(L)=\lim_{x\to\infty}\left(\frac{\ln\left(e^x+x\right)}{x}\right)$$

This is the indeterminate form $$\frac{\infty}{\infty}$$ and so we may apply L'Hôpital's Rule.

Can you proceed?
 
Is the final answer of the limit e ? I have proceeded and got that the right hand side is 1.

What do you mean to derive implicitly?
 
Yankel said:
Is the final answer of the limit e ? I have proceeded and got that the right hand side is 1.

Yes, I got the same result. :D

Yankel said:
What do you mean to derive implicitly?

Implicit differentiation is nothing more than a special case of the well-known chain rule for derivatives. The majority of differentiation problems in first-year calculus involve functions $y$ written EXPLICITLY as functions of $x$. However, some functions $y$ are written IMPLICITLY as functions of $x$, as is the case once we took the natural logs of both sides. So, we differentiate, bearing in mind that the chain rule must be used with $y$ since it is a function of $x$. :D
 

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