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I tried one approach but the correction in the book shows me a total different answer.

Here's what I did:

$(1+ 1/x)^x=xln(1+1/x)$ Thus, now we try to find the derivative of a multiplication:

$ u(x)=x$

$(u(x))'=1$

$v(x)=ln(1+1/x)$

$(v(x))'= -1/(x^2) +1/x$

And so:

$(uv)'=u'v+uv'$ which gives us:

$(uv)'=xln(1 +1/x) +(-1/x^2 -1/x)$

Yet, the correction gives me this as an answer:

$ln(1+1/x)-1/(1+x)$

John Mayne
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2 Answers2

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Let $f(x) = (1+1/x)^x$. Then $$\ln(f(x)) = x \ln(1 + 1/x) = x \ln( (x+1)/x) = x \ln(x+1) - x \ln(x).$$

Taking derivatives on both sides gives:

$$\frac{f'(x)}{f(x)} = \ln(x+1) + \frac{x}{x+1} - \ln(x) - 1.$$

We wish to find $f'(x)$, so we solve for it and subsitute the value of $f(x)$.

$$f'(x) = \left(1 + \frac{1}{x} \right)^x \left[ \ln(x+1) + \frac{x}{x+1} - \ln(x) - 1 \right]=\left(1 + \frac{1}{x} \right)^x \left[ \ln\left( 1 + \frac{1}{x} \right) - \frac{1}{x+1} \right]$$

Joel
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  • I don't understand how you go from finding what $ln(f(x))$ equals to, to what $f'(x)/f(x)$ equals to. Can you explain? I can see how you use $f'(x)/f(x)$ to find $f'(x)$ afterwards. – John Mayne Jul 18 '16 at 19:50
  • When you take the derivative of $\ln(f(x))$ you must use the chain rule. Thus $$\frac{d}{dx} \ln(f(x)) = \frac{1}{f(x)} \cdot f'(x).$$ – Joel Jul 18 '16 at 20:13
  • I still don't understand how you can find what $f'(x)/f(x)$ equals to without knowing what $f'(x)$ equals to. – John Mayne Jul 18 '16 at 20:30
  • What we know is that $\ln(f(x)) = x \ln(1+1/x)$. We can certainly take the derivative of the right hand side. The derivative of the left hand side must agree with that of the right, and the derivative of the left is $f'(x)/f(x)$. – Joel Jul 18 '16 at 20:58
  • At that point we don't know $f'(x)$, we only know the ratio $f'(x)/f(x)$. From there we solve for $f'(x)$. It's the same if we are trying to find some unknown $z$, but we only know $z/y= 5$ and $y=3$. $f'(x)$ is an unknown that we are trying to solve for. – Joel Jul 18 '16 at 21:00
  • This must sound really dumb, but how do we know the ration $f'(x)/f(x)$ ? – John Mayne Jul 18 '16 at 21:26
  • We know that $f'(x)/f(x)$ is the derivative of $\ln(f(x))$, by the chain rule. Since we know the equation for $\ln(f(x))$ is $x \ln(1+1/x)$, we can compute another form for the derivative of $\ln(f(x))$ by taking the derivative of $x\ln(1+1/x)$. These must be the same, since they are both the derivative of $\ln(f(x))$. – Joel Jul 18 '16 at 21:36
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$y = (1 + 1/x)^x\\ \ln y = x \ln (1+1/x)\\ \frac 1y y' = \ln(1+1/x) + x(\frac 1{1+\frac1x})(-\frac 1{x^2})\\ y' = (1 + 1/x)^x (\ln(1+1/x) - \frac 1{x+1})\\ y' = (1 + 1/x)^x (\ln(x+1) - \ln x - \frac 1{x+1})$

Doug M
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