What is the first derivative of this function?

  • 1
    $\begingroup$ @Amzoti - I don't think the power rule work... $\endgroup$
    – Victor
    Commented Apr 15, 2013 at 1:15
  • 2
    $\begingroup$ It is Logarithmic differentiation (derivative), see youtube.com/watch?v=x6j9khJSArw $\endgroup$
    – Amzoti
    Commented Apr 15, 2013 at 1:19

2 Answers 2


You will find what you are looking for here. The method described there is known as logarithmic differentiation. It applies very well here since you have a composition of functions of $n$.

image from the wikipedia page: https://i.sstatic.net/2nZOE.png


Since $\sqrt[n]{n}=n^{1/n}=e^{(1/n)\ln{n}}$, you can find the derivative of that:

$$\frac{\mathrm{d}}{\mathrm{d}n} e^{(1/n)\ln{n}}$$

Let $u = \frac{\ln{n}}{n}$ and use the chain rule,

$$\frac{\mathrm{d}}{\mathrm{d}n} \sqrt[n]{n} = \frac{\mathrm{d}}{\mathrm{d}n} e^{(1/n)\ln{n}} = \frac{\mathrm{d}}{\mathrm{d}n} e^{u} = e^u \cdot \frac{\mathrm{d}u}{\mathrm{d}n}$$

Use the division rule:

$$f = \frac{g}{h} \implies f^\prime = \frac{g^\prime h - g h^\prime}{h^2}$$

The derivative of $\ln{n}$ is $1/n$, and the derivative of $n$ is $1$ so:

$$\frac{\mathrm{d}}{\mathrm{d}n} \sqrt[n]{n} = e^u \frac{\frac{1}{n} n - \ln{n} \cdot 1}{n^2} = e^u \frac{1 - \ln{n}}{n^2}$$

Substitute back in for $u$:

$$\frac{\mathrm{d}}{\mathrm{d}n} \sqrt[n]{n} = e^{(1/n) \ln{n}} \frac{1 - \ln{n}}{n^2} = \sqrt[n]{n} \frac{1 - \ln{n}}{n^2}$$

  • $\begingroup$ Welcome to MSE! It helps to format answers using MathJax (see FAQ). regards $\endgroup$
    – Amzoti
    Commented Apr 15, 2013 at 1:26
  • $\begingroup$ this is a very clever (albeit elaborate) way to arrive at the solution, but see the link in my answer which describes a very powerful method for dealing with this problem, as well as many other similar problems in a more lightweight manner. $\endgroup$ Commented Apr 15, 2013 at 4:27
  • $\begingroup$ Cool, and thanks for putting my answer into mathjax, I kind of didn't know about that. $\endgroup$
    – hacatu
    Commented Apr 15, 2013 at 19:49

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