Differentiate from first principle
1.
(We know that if
then
)
2.
![{\displaystyle f(z)=(1-z)^{2}=z^{2}-2z+1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3c1137aa52fcdd7553468c2a18786290f00be9ac)
![{\displaystyle f'(z)=2z-2}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9aebcd7789b7cfd4a43afb028e50b0d9bca67c59)
3.
![{\displaystyle {\begin{matrix}f'(z)&=&\lim _{h\to 0}{\frac {{\frac {1}{(1-z-h)^{2}}}-{\frac {1}{(1-z)^{2}}}}{h}}\\&=&\lim _{h\to 0}{\frac {1}{h}}({\frac {1}{(1-z-h)^{2}}}-{\frac {1}{(1-z)^{2}}})\\&=&\lim _{h\to 0}{\frac {1}{h}}({\frac {(1-z)^{2}}{(1-z-h)^{2}(1-z)^{2}}}-{\frac {(1-z-h)^{2}}{(1-z-h)^{2}(1-z)^{2}}})\\&=&\lim _{h\to 0}{\frac {1}{h}}{\frac {(1-z)^{2}-(1-z-h)^{2}}{(1-z-h)^{2}(1-z)^{2}}}\\&=&\lim _{h\to 0}{\frac {1}{h}}{\frac {z^{2}-2z+1-(z^{2}+2hz-2z+h^{2}-2h+1)}{(1-z-h)^{2}(1-z)^{2}}}\\&=&\lim _{h\to 0}{\frac {1}{h}}{\frac {z^{2}-2z+1-z^{2}-2hz+2z-h^{2}+2h-1)}{(1-z-h)^{2}(1-z)^{2}}}\\&=&\lim _{h\to 0}{\frac {1}{h}}{\frac {-2hz-h^{2}+2h}{(1-z-h)^{2}(1-z)^{2}}}\\&=&\lim _{h\to 0}{\frac {-2z-h+2}{(1-z-h)^{2}(1-z)^{2}}}\\&=&{\frac {-2z+2}{(1-z)^{2}(1-z)^{2}}}\\&=&{\frac {-2z+2}{(1-z)^{4}}}\\&=&{\frac {2(1-z)}{(1-z)^{4}}}\\&=&{\frac {2}{(1-z)^{3}}}\\\end{matrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/866bf30a8f6520921ccb9d6c83fc7181e6627b58)
4.
![{\displaystyle f(z)=(1-z)^{3}=-z^{3}+3z^{2}-3z+1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4ae1705c203f835fb6dbe18cf3f62a8ed26cfcc9)
![{\displaystyle f'(z)=-3z^{2}+6z-3}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2ad15b91048ef188d665de00412d200a45d258d3)
5. if
- f(x)=g(x)+h(x)
then
Differentiating f(z) = (1 - z)^n
1.
![{\displaystyle f(z)=(1-z)^{3}=-z^{3}+3z^{2}-3z+1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4ae1705c203f835fb6dbe18cf3f62a8ed26cfcc9)
![{\displaystyle {\begin{matrix}f'(z)&=&-3z^{2}+6z-3\\&=&-3(z^{2}-2z+1)\\&=&-3(z-1)^{2}\\\end{matrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7d6dfe004813b95f0018bf84856a3c32962dc28e)
2.
![{\displaystyle f(z)=(1+z)^{2}=z^{2}+2z+1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/31c3a75fc018726db39fbd3eddf528bcf0186f60)
![{\displaystyle {\begin{matrix}f'(z)&=&2z+2\\&=&2(z+1)\end{matrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/97c78397d25f091410ad237fa532030911f8d461)
3.
![{\displaystyle f(z)=(1+z)^{3}=z^{3}+3z^{2}+3z+1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0710047734b7667ee3bc4b9abefda6ab32c50ee5)
![{\displaystyle {\begin{matrix}f'(z)&=&3z^{2}+6z+3\\&=&3(z^{2}+2z+1)\\&=&3(z+1)^{2}\\\end{matrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c42563bf043bd223148ae5e209b5e89d706311ba)
4.
![{\displaystyle f(z)={\frac {1}{(1-z)^{3}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f072e5154633971b9e6c8577b5cc21806761ebe5)
![{\displaystyle {\begin{matrix}f'(z)&=&\lim _{k\to 0}{\frac {{\frac {1}{(1-z-k)^{3}}}-{\frac {1}{(1-z)^{3}}}}{k}}\\&=&\lim _{k\to 0}{\frac {1}{k}}({\frac {1}{(1-z-k)^{3}}}-{\frac {1}{(1-z)^{3}}})\\&=&\lim _{k\to 0}{\frac {1}{k}}{\frac {(1-z)^{3}-(1-z-k)^{3}}{(1-z-k)^{3}(1-z)^{3}}}\\&=&\lim _{k\to 0}{\frac {1}{k}}{\frac {-z^{3}+3z^{2}-3z+1-(-z^{3}-3kz^{2}+3z^{2}-3k^{2}z+6kz-3z-k^{3}+3k^{2}-3k+1)}{(1-z-k)^{3}(1-z)^{3}}}\\&=&\lim _{k\to 0}{\frac {1}{k}}{\frac {-z^{3}+3z^{2}-3z+1+z^{3}+3kz^{2}-3z^{2}+3k^{2}z-6kz+3z+k^{3}-3k^{2}+3k-1}{(1-z-k)^{3}(1-z)^{3}}}\\&=&\lim _{k\to 0}{\frac {1}{k}}{\frac {3kz^{2}+3k^{2}z-6kz+k^{3}-3k^{2}+3k)}{(1-z-k)^{3}(1-z)^{3}}}\\&=&\lim _{k\to 0}{\frac {3z^{2}+3kz-6z+k^{2}-3k+3)}{(1-z-k)^{3}(1-z)^{3}}}\\&=&{\frac {3z^{2}-6z+3}{(1-z)^{3}(1-z)^{3}}}\\&=&{\frac {3z^{2}-6z+3}{(1-z)^{6}}}\\&=&{\frac {3(z^{2}-2z+1)}{(1-z)^{6}}}\\&=&{\frac {3(1-z)^{2}}{(1-z)^{6}}}\\&=&{\frac {3}{(1-z)^{4}}}\\\end{matrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2fdeef2851a50e6281f2605dbcfcbef247c832a3)
Differentiation technique
1.
![{\displaystyle {\begin{matrix}f(z)&=&{\frac {1}{(1-z)^{2}}}\\f'(z)&=&-{\frac {((1-z)^{2})'}{((1-z)^{2})^{2}}}\\&=&-{\frac {-2(1-z)}{(1-z)^{4}}}\\&=&{\frac {2}{(1-z)^{3}}}\end{matrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a4bc5b8ad5bb46bbcb561e65f32683c63dbab7a3)
We use the result of the differentation of f(z)=(1-z)^n (f'(z) = -n(1-z)n-1)
2.
![{\displaystyle {\begin{matrix}f(z)&=&{\frac {1}{(1-z)^{3}}}\\f'(z)&=&-{\frac {((1-z)^{3})\prime }{((1-z)^{3})^{2}}}\\&=&-{\frac {-3(1-z)^{2}}{(1-z)^{6}}}\\&=&{\frac {3}{(1-z)^{4}}}\end{matrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/43e524b3446e3101de00bd99cc92fa1513902f2c)
3.
![{\displaystyle {\begin{matrix}f(z)&=&{\frac {1}{(1+z)^{3}}}\\f'(z)&=&-{\frac {((1+z)^{3})'}{((1+z)^{3})^{2}}}\\&=&-{\frac {3(1+z)^{2}}{(1+z)^{6}}}\\&=&{\frac {-3}{(1+z)^{4}}}\end{matrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3254e9d0c86537fa3f3e7041c3d0c2833afcc6d1)
We use the result of exercise 3 of the previous section f(z)= (1+z)3 -> f'(z)=3(1+z)^2
4.
![{\displaystyle {\begin{matrix}f(z)&=&{\frac {1}{(1-z)^{n}}}\\f'(z)&=&-{\frac {((1-z)^{n})'}{((1-z)^{n})^{2}}}\\&=&-{\frac {-n(1-z)^{n-1}}{(1-z)^{2n}}}\\&=&-{\frac {-n}{(1-z)^{2n-(n-1)}}}\\&=&-{\frac {-n}{(1-z)^{2n-n+1}}}\\&=&{\frac {n}{(1-z)^{n+1}}}\end{matrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b611b18b2a7548d73bff0b62339ba5fb7c78acac)
We use the result of the differentation of f(z)=(1-z)^n (f'(z) = -n(1-z)n-1)