Partial derivative of $l(\mu,\lambda;\underline{y})=\frac{n}{2}\log\lambda-\frac{\lambda}{2\mu^2}(n\overline{y}-2n\mu+\mu^2\sum^n_{i=1}\frac{1}{y_i})$

Question

For a part of one of my exercises, I have to derive

$$ \frac{\partial}{\partial \mu}l(\mu,\lambda;\underline{y}) $$

where $ l(\mu,\lambda;\underline{y})=\frac{n}{2}\log\lambda-\frac{\lambda}{2\mu^2}(n\overline{y}-2n\mu+\mu^2\sum^n_{i=1}\frac{1}{y_i})$ and $\underline{y}$ is a set $(y_1,y_2,...,y_n)$.

What I have gotten:

$$ \frac{\partial}{\partial \mu}l(\mu,\lambda;\underline{y}) \\ = -\frac{\lambda}{2\mu^2}(-2n+2\mu\sum^n_{i=1}\frac{1}{y_i})-\frac{\lambda}{2}\cdot(-2)\cdot\frac{1}{\mu^3}\cdot(n\overline{y}-2n\mu+\mu^2\sum^n_{i=1}\frac{1}{y_i}) \\ = \frac{\lambda}{\mu^3}(n\overline{y}-2n\mu+\mu^2\sum^n_{i=1}\frac{1}{y_i})-\frac{\lambda}{2\mu^2}(-2n+2\mu\sum^n_{i=1}\frac{1}{y_i}).$$

The right answer should be

$$\frac{n\lambda}{\mu^3}(\overline{y}-\mu)$$

but I don't know how to get to this.

Answer 1

$$\require{cancel}\frac{\lambda}{\mu^3}\left(n\overline{y}-2n\mu+\mu^2\sum^n_{i=1}\frac{1}{y_i}\right)-\frac{\lambda}{\cancel2\mu^2}\left(-\cancel{2}n+\cancel2\mu\sum^n_{i=1}\frac{1}{y_i}\right)=$$

$$=\frac\lambda{\mu^3}\left[n\overline y-\color{red}{2n\mu}+\cancel{\mu^2\sum_{i=1}^n\frac1{y_i}}+\color{red}{n\mu-}\cancel{\mu^2\sum_{i=1}^n\frac1{y_i}}\right]=$$

$$=\frac\lambda{\mu^3}\left[n\overline y-\color{red}{n\mu}\right]=\frac{n\lambda}{\mu^3}\left(\overline y-\mu\right)$$