$e^{-1/2 \sum_{i=1}^n (x_i - \theta)^2}$ wrt to $\theta$? (Without log.)
How to differentiate a part of normal likelihood function
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0Oh I mean product rule above. – 2012-10-30
1 Answers
Well, if that helps you, using the chain rule:
$\left(e^{-\frac{1}{2}\sum^{n}_{i=1}(x_{i}-\theta)^2}\right)^{\prime}=e^{-\frac{1}{2}\sum^{n}_{i=1}(x_{i}-\theta)^2}\cdot\left(-\frac{1}{2}\right)\sum^{n}_{i=1}(2\theta-2x_{i})=-e^{-\frac{1}{2}\sum^{n}_{i=1}(x_{i}-\theta)^2}\sum^{n}_{i=1}(\theta-x_{i})$
EDIT
To respond to your comment - if want to equate this expression to $0$, we know that $\sum^n_{i=1}(\theta-x_{i})=0$ because the exponential function does not reach zero for any argument.
EDIT 2
To illustrate how the chain rule works, let $f(\theta)=-\frac{1}{2}(x_{i}-\theta)^2$ and $h(\theta)=e^{f(\theta)}$. The derivative of $h$ is thus equal to:
$h^{\prime}(\theta)=h(\theta)\cdot f^{\prime}(\theta)$
We know that $f=-\frac{1}{2}(x_{i}^{2}-2x_{i}\theta+\theta^{2})=-\frac{1}{2}x_{i}+x_{i}\theta-\frac{\theta^{2}}{2}$, hence $f^{\prime}(\theta)=x_{i}-\theta=-\frac{1}{2}(2\theta-2x_{i})$
Thus finally:
$h^{\prime}(\theta)=h(\theta)\cdot f^{\prime}(\theta)=e^{-\frac{1}{2}(x_{i}-\theta)^2}\cdot -\frac{1}{2}(2\theta-2x_{i})=-e^{-\frac{1}{2}(x_{i}-\theta)^2}\cdot(\theta-x_{i})$
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0If you don't have any further questions, would you please consider accepting my answer? – 2012-10-30