I am attempting a problem from Larsen and Marx, 4th edition that asks to find the maximum likelihood estimate for $\theta$ in the pdf: $ f(y; \theta) = \dfrac{2y}{1-\theta^{2}}, \theta \leq y \leq 1$
It also states that a random sample of size 6 yielded measurements 0.70, 0.63, 0.92, 0.86, 0.43, and 0.21. I used the definition of the likelihood function to get: $L(\theta) = \prod_{i=1}^n \dfrac{2y_{i}}{1-\theta^{2}} = 2^{n}(1-\theta^{2})^{-n} * \prod_{i=1}^n y_{i}$
Because it is easier to deal with $\ln L(\theta)$ for the purpose of deriving and finding a $\theta$ that maximizes $L(\theta)$: $\ln L(\theta) = -n \ln (1-\theta^{2}) + n \ln 2 + \ln \prod_{i=1}^n y_{i}$ $\dfrac{d \ln L(\theta)}{d\theta} = \dfrac{2n\theta}{(1-\theta^{2})}$
However, setting this derivative to zero would mean $\theta_{e} = 0$ regardless of sample size and leaves me with a pdf of $f_{Y} = 2y$. Did I go wrong somewhere in my solution?