Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I need to differentiate

$$-\frac{v}{2}\ln(2) - \ln\,\Gamma\left(\frac{v}{2}\right)$$

I did it and got

$$-\frac{\ln(2)}{2} - \frac{\Gamma'(\frac{v}{2})}{\Gamma(\frac{v}{2})}$$

But the answers say it should be

$$-\frac{\ln(2)}{2} - \frac{\Gamma'(\frac{v}{2})}{2\Gamma(\frac{v}{2})}$$

How?

share|improve this question
    
For some basic information about writing math at this site see e.g. here, here, here and here. –  Américo Tavares Jan 15 '13 at 15:41

2 Answers 2

up vote 3 down vote accepted

$$\frac{d}{dv}\left(-\frac{v}{2}\ln 2 -\ln \Gamma \left(\frac{v}{2}\right)\right)=-\frac{1}{2}\ln 2 -\frac{1}{\Gamma(v/2)} \frac{d}{dv} \Gamma\left(\frac{v}{2}\right)\ =$$

$$= -\frac{1}{2}\ln 2 -\frac{1}{\Gamma(v/2)}\cdot \frac{1}{2} \Gamma '\left(\frac{v}{2}\right)\ $$

share|improve this answer
1  
So I have to use chain rule to do $\frac{d}{dv} \left( \Gamma \left(\frac{v}{2} \right) \right)$? –  Kaish Jan 15 '13 at 15:06
2  
Of course, @Kaish: how else? It is the composition of two functions.\ –  DonAntonio Jan 15 '13 at 15:10

Let's $f(u)=\frac{u}{2}$. Then, $$ -\frac{v}{2}\ln(2) - \ln\,\Gamma\left(\frac{v}{2}\right)= -\ln(2)\cdot f(u)-\ln\circ \Gamma\circ f(u) $$ By chain rule, \begin{align} \frac{d}{du}\bigg[-\ln(2)\cdot f(u)-\ln\circ \Gamma\circ f(u)\bigg]= & -\ln(2)\cdot \frac{1}{2} \\ -& \bigg( D_z\ln(z)\bigg|_{z=\Gamma\circ f(u)} \bigg) \cdot \bigg(D_y\Gamma(y)\bigg|_{y=f(u)} \bigg)\cdot \bigg( D_x f(x)\bigg|_{x=u} \bigg) \end{align}

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.