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I am having a problem differentiating. Thinking about the chain rule, but not sure how to get this to work.

Would like to calculate

$$\frac{\partial}{\partial\mu}\Bigg[\Bigg(\frac{p_0}{p_b(\mu)}\Bigg)^{\beta(\mu)}\Bigg]$$

Any help will be much appreciated.

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    $\begingroup$ Convert it to exponential and logarithm functions, like $a^b = \exp(b \log a)$. Then you can apply the chain rule and the product rule as usual. $\endgroup$ – Tunococ Jan 30 '13 at 11:26
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Hint:

I assume that $p_0$ and $b$ are constants. Set $y=\left(\frac{p_0}{p_b(\mu)}\right)^{\beta{(\mu)}}$ and by taking logarithym from both sides we have $$\ln(y)=\beta{(\mu)}\ln\left(\frac{p_0}{p_b(\mu)}\right)$$ Now differentiate form both sides respect to $\mu$: $$\frac{\partial}{\partial\mu}\ln(y)=\frac{\partial}{\partial\mu}\left(\beta{(\mu)}\ln\left(\frac{p_0}{p_b(\mu)}\right)\right)$$ so $$\frac{\partial y}{y\partial\mu}=\frac{\partial}{\partial\mu}(\beta{(\mu)})\ln\left(\frac{p_0}{p_b(\mu)}\right)+\frac{\partial}{\partial\mu}\left(\ln\left(\frac{p_0}{p_b(\mu)}\right)\right)(\beta{(\mu)})$$ so

$$\frac{\partial y}{\partial\mu}=y\left[\frac{\partial\beta}{\partial\mu}\ln\left(\frac{p_0}{p_b(\mu)}\right)-\frac{\partial{p_b}}{\partial\mu}\frac{\beta{(\mu)}}{p_b(\mu)}\right]$$

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    $\begingroup$ +1 Nice...in particular the stamina required to write all the time the code for "mu" instead of changing the whole thing altogether to "u". $\endgroup$ – DonAntonio Jan 30 '13 at 12:09
  • $\begingroup$ @DonAntonio: Thanks Don for the time you edited mine. :D $\endgroup$ – mrs Jan 30 '13 at 12:16
  • $\begingroup$ @BabakSorouh Thank you $\endgroup$ – Bastaix Jan 30 '13 at 12:19
  • $\begingroup$ I inly hope that my answer made you way smooth. $\endgroup$ – mrs Jan 30 '13 at 12:20
  • $\begingroup$ @Bastaix: I assumed $p_0$ is a constant, otherwise, my answer might be incorrect in its last terms. If we accept $p_0$ is a constant so $\ln(p_0/p_b(\mu))=\ln(p_0)-\ln(p_b(\mu))$. The minus caused here when we differentiate it. $\endgroup$ – mrs Jan 30 '13 at 12:31

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