1 | ||
f(t) = | ||
√1−2xt+t2 |
dnf | ||
Skonstruować pętle pozwalające obliczyć | |t=0 | |
dtn |
1 | ||
f(t) = | ||
√1−2xt+t2 |
1 | −2x+2t | |||
f'(t) =(− | ) | |||
2 | (1−2xt+t2)3/2 |
x−t | ||
f'(t) = | ||
(1−2xt+t2)3/2 |
−1 | 3 | (−2x+2t)(x−t) | ||||
f''(t) = | + (− | ) | ||||
(1−2xt+t2)3/2 | 2 | (1−2xt+t2)(5/2) |
−1 | 3(x−t)2 | |||
f''(t) = | + | |||
(1−2xt+t2)3/2 | (1−2xt+t2)5/2 |
−3 | (−2x+2t) | |||
f'''(t) = (−1)( | ) | +U{3*2*(x−t)*(−1)}{(1−2xt+t2)5/2 | ||
2 | (1−2xt+t2)5/2 |
5 | (−2x+2t)*3(x−t)2 | |||
+(− | ) | |||
2 | (1−2xt+t2)7/2 |
−9(x−t) | 15(x−t)3 | |||
f'''(t) = | + | |||
(1−2xt+t2)5/2 | (1−2xt+t2)7/2 |
−9*(−1) | 5 | (−2x+2t)(−9(x−t)) | ||||
f(4)(t) = | +(− | ) | ||||
(1−2xt+t2)5/2 | 2 | (1−2xt+t2)7/2 |
15*3*(x−t)2*(−1) | ||
+ | + | |
(1−2xt+t2)7/2 |
7 | (−2x+2t)(15(x−t)3) | |||
(− | ) | |||
2 | (1−2xt+t2)9/2 |
9 | 90(x−t)2 | |||
f(4)(t) = | − | + | ||
(1−2xt+t2)5/2 | (1−2xt+t2)7/2 |
105(x−t)4 | ||
(1−2xt+t2)9/2 |