The second-derivative of the log-likelihood function in logistic regression (page 120) model In textbook The Elements of Statistical Learning, for logistic regression (page 120) model, the log-likelihood function can be written as
$$\ell(\beta)=\sum_{i=1}^N(y_i\beta^Tx_i-\log(1+e^{\beta^Tx_i}))$$
This is a $(p+1)$ nonlinear equations in $\beta$. To solve this problem, we use the Newton-Raphson algorithm, which requires the second-derivative
$$\frac{\partial^2 \ell(\beta)}{\partial \beta \partial \beta^T}=-\sum_{i=1}^Nx_ix_i^Tp(x_i;\beta)(1-p(x_i;\beta))$$
where $p(x_i;\beta)=1-\frac{1}{1+e^{\beta^Tx_i}}$.
My questions are
(1) why the second derivate is not defined by
$$\frac{\partial^2 \ell(\beta)}{\partial \beta ^2}?$$
(2) What is the result of
$$\frac{\partial \ell(\beta)}{\partial \beta^T}=\frac{e^{\beta^Tx_i}x_i}{(1+e^{\beta^Tx_i})^2}?$$
Is it same as $\frac{\partial \ell(\beta)}{\partial \beta}?$

Maybe you are confused by the difference between univariate and multivariate differentiation.
Your first derivative is wrt to a vector $\boldsymbol{\beta}$ and therefore is expected to be a vector itself (the collection of all partial derivatives). I.e. $$\frac{\partial \mathcal{l}}{\partial \boldsymbol{\beta}^T}= \left[\frac{\partial \mathcal{l}}{\partial \beta_0},\ldots,\frac{\partial \mathcal{l}}{\partial \beta_p}\right]^T.$$
Now, the second derivative is formed by deriving every element  in $\frac{\partial \mathcal{l}}{\partial \boldsymbol{\beta}^T}$ wrt to $\boldsymbol{\beta}$ again, so we have to form $p +1$ derivatives for each of the $p +1$ elements in $\frac{\partial \mathcal{l}}{\partial \boldsymbol{\beta}^T}$, which results in a $(p +1) \times (p +1)$ matrix, the Hessian.
So to answer your first question, you have $\partial \beta^2$ in the denominator when you derive wrt to one coefficient (the univariate version) and you have $\partial \boldsymbol{\beta}^T\boldsymbol{\beta}$ in the denominator when you derive wrt to a vector (the multivariate version).
Your second question is related to the first question. The result of $\frac{\partial \mathcal{l}}{\partial \boldsymbol{\beta}^T}$ is a vector and is therefore not the same as the result of $\frac{\partial \mathcal{l}}{\partial\beta}$ which is a scalar.
