dstefan
03-22-2008, 01:12 PM
Chapter 1, Problem 8:
Let f(y) = \sum_{i=1}^n c_i e^{-y t_i}, where c_i and t_i, i=1:n, are positive constants. Compute f'(y) and f''(y).
Solution:
Note that
\left( e^{-y t_i} \right)^{'} ~=~ \frac{d}{dy} \left( e^{-y t_i} \right) ~=~ - t_i e^{-y t_i}
\left( e^{-y t_i} \right)^{''} ~=~ \frac{d}{dy} \left( - t_i e^{-y t_i} \right) ~=~ t_i^2 e^{-y t_i}
Then,
f'(y) ~=~ - \sum_{i=1}^n c_i t_i e^{-y t_i}
f^{''}(y) ~=~ \sum_{i=1}^n c_i t_i^2 e^{-y t_i}
Let f(y) = \sum_{i=1}^n c_i e^{-y t_i}, where c_i and t_i, i=1:n, are positive constants. Compute f'(y) and f''(y).
Solution:
Note that
\left( e^{-y t_i} \right)^{'} ~=~ \frac{d}{dy} \left( e^{-y t_i} \right) ~=~ - t_i e^{-y t_i}
\left( e^{-y t_i} \right)^{''} ~=~ \frac{d}{dy} \left( - t_i e^{-y t_i} \right) ~=~ t_i^2 e^{-y t_i}
Then,
f'(y) ~=~ - \sum_{i=1}^n c_i t_i e^{-y t_i}
f^{''}(y) ~=~ \sum_{i=1}^n c_i t_i^2 e^{-y t_i}