Estimating the parameter $\beta$ The lifetime of computer monitors has a exponential distribution where the expected value can be written as:
$\mu(s) = \frac{\beta}{s}$
Where $s$ is how bright the monitor is and both $s$ and $\beta$ are >0
How do I estimate the maximum-likelihood of parameter $\beta$?
I know that the log-likelihood for a exponential distribution is:
$l(\lambda) = n(\ln(\lambda) - \lambda\bar{t})$
And taking the derivative and equals it to zero yields:
$\frac{dl(\lambda)}{d\lambda} = n(\frac{1}{\lambda} - \bar{t}) = 0$
$\bar{t} = \frac{1}{\lambda}$
Here is where I get stuck.
EDIT 1
I have data for both the lifetime of the computer monitors and their brightness.

Assise from using a direct derivation, leading to a closed form expression, you can do this with a generalized linear model a description is given in the question
Fitting exponential (regression) model by MLE?
With that approach you have more flexibility (e.g. change the function $\mu(s)$)
Example
In R you would use the function glm with the family gamma and specify a dispersion parameter dispersion=1 (the exponential distribution is a special case of the gamma distribution for which the dispersion parameter is equal to 1).
### generate and plot data 

set.seed(1)
n = 100
x = runif(n,0,10)
mu = 2/x
rate = 1/mu
y = rexp(n,rate)
plot(x,y)

#### fit glm model with Gamma distribution that has k=1 (ie. the exponential distribution)

fit = glm(y ~ 0+x, family = Gamma())
summary(fit,dispersion=1)

# This gives as result 
# a rate parameter of lambda = 0.49868 x
# which relates to
# a mean that is mu = 1/lambda = 2.005294/x

