High School Mathematics Extensions/Supplementary/Differentiation/Solutions

High School Mathematics Extensions

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Supplementary Chapters — Primes and Modular Arithmetic — Logic

Mathematical Proofs — Set Theory and Infinite Processes — Counting and Generating Functions — Discrete Probability

Matrices — Further Modular Arithmetic — Mathematical Programming — Markov Chains

Differentiate from first principle

1. $f'(z)=3z^{2}$ (We know that if $p(x)=x^{n}$ then $p'(x)=nx^{n-1}$)

2.

$f(z)=(1-z)^{2}=z^{2}-2z+1$

$f'(z)=2z-2$

3.

${\begin{matrix}f'(z)&=&\lim _{h\to 0}{\frac {{\frac {1}{(1-z-h)^{2}}}-{\frac {1}{(1-z)^{2}}}}{h}}\\&=&\lim _{h\to 0}{\frac {1}{h}}({\frac {1}{(1-z-h)^{2}}}-{\frac {1}{(1-z)^{2}}})\\&=&\lim _{h\to 0}{\frac {1}{h}}({\frac {(1-z)^{2}}{(1-z-h)^{2}(1-z)^{2}}}-{\frac {(1-z-h)^{2}}{(1-z-h)^{2}(1-z)^{2}}})\\&=&\lim _{h\to 0}{\frac {1}{h}}{\frac {(1-z)^{2}-(1-z-h)^{2}}{(1-z-h)^{2}(1-z)^{2}}}\\&=&\lim _{h\to 0}{\frac {1}{h}}{\frac {z^{2}-2z+1-(z^{2}+2hz-2z+h^{2}-2h+1)}{(1-z-h)^{2}(1-z)^{2}}}\\&=&\lim _{h\to 0}{\frac {1}{h}}{\frac {z^{2}-2z+1-z^{2}-2hz+2z-h^{2}+2h-1)}{(1-z-h)^{2}(1-z)^{2}}}\\&=&\lim _{h\to 0}{\frac {1}{h}}{\frac {-2hz-h^{2}+2h}{(1-z-h)^{2}(1-z)^{2}}}\\&=&\lim _{h\to 0}{\frac {-2z-h+2}{(1-z-h)^{2}(1-z)^{2}}}\\&=&{\frac {-2z+2}{(1-z)^{2}(1-z)^{2}}}\\&=&{\frac {-2z+2}{(1-z)^{4}}}\\&=&{\frac {2(1-z)}{(1-z)^{4}}}\\&=&{\frac {2}{(1-z)^{3}}}\\\end{matrix}}$

4.

$f(z)=(1-z)^{3}=-z^{3}+3z^{2}-3z+1$

$f'(z)=-3z^{2}+6z-3$

5. if

f(x)=g(x)+h(x)

then

${\begin{matrix}f'(x)&=&\lim _{k\to 0}{\frac {f(x+k)-f(x)}{k}}\\&=&\lim _{k\to 0}{\frac {(g(x+k)+h(x+k))-(g(x)+h(x))}{k}}\\&=&\lim _{k\to 0}{\frac {g(x+k)-g(x)+h(x+k)-h(x)}{k}}\\&=&\lim _{k\to 0}({\frac {g(x+k)-g(x)}{h}}+{\frac {h(x+k)-h(x)}{k}})\\&=&\lim _{k\to 0}{\frac {g(x+k)-g(x)}{h}}+\lim _{k\to 0}{\frac {h(x+k)-h(x)}{k}}\\&=&g'(x)+h'(x)\\\end{matrix}}$

Differentiating f(z) = (1 - z)^n

1.

$f(z)=(1-z)^{3}=-z^{3}+3z^{2}-3z+1$

${\begin{matrix}f'(z)&=&-3z^{2}+6z-3\\&=&-3(z^{2}-2z+1)\\&=&-3(z-1)^{2}\\\end{matrix}}$

2.

$f(z)=(1+z)^{2}=z^{2}+2z+1$

${\begin{matrix}f'(z)&=&2z+2\\&=&2(z+1)\end{matrix}}$

3.

$f(z)=(1+z)^{3}=z^{3}+3z^{2}+3z+1$

${\begin{matrix}f'(z)&=&3z^{2}+6z+3\\&=&3(z^{2}+2z+1)\\&=&3(z+1)^{2}\\\end{matrix}}$

4.

$f(z)={\frac {1}{(1-z)^{3}}}$

${\begin{matrix}f'(z)&=&\lim _{k\to 0}{\frac {{\frac {1}{(1-z-k)^{3}}}-{\frac {1}{(1-z)^{3}}}}{k}}\\&=&\lim _{k\to 0}{\frac {1}{k}}({\frac {1}{(1-z-k)^{3}}}-{\frac {1}{(1-z)^{3}}})\\&=&\lim _{k\to 0}{\frac {1}{k}}{\frac {(1-z)^{3}-(1-z-k)^{3}}{(1-z-k)^{3}(1-z)^{3}}}\\&=&\lim _{k\to 0}{\frac {1}{k}}{\frac {-z^{3}+3z^{2}-3z+1-(-z^{3}-3kz^{2}+3z^{2}-3k^{2}z+6kz-3z-k^{3}+3k^{2}-3k+1)}{(1-z-k)^{3}(1-z)^{3}}}\\&=&\lim _{k\to 0}{\frac {1}{k}}{\frac {-z^{3}+3z^{2}-3z+1+z^{3}+3kz^{2}-3z^{2}+3k^{2}z-6kz+3z+k^{3}-3k^{2}+3k-1}{(1-z-k)^{3}(1-z)^{3}}}\\&=&\lim _{k\to 0}{\frac {1}{k}}{\frac {3kz^{2}+3k^{2}z-6kz+k^{3}-3k^{2}+3k)}{(1-z-k)^{3}(1-z)^{3}}}\\&=&\lim _{k\to 0}{\frac {3z^{2}+3kz-6z+k^{2}-3k+3)}{(1-z-k)^{3}(1-z)^{3}}}\\&=&{\frac {3z^{2}-6z+3}{(1-z)^{3}(1-z)^{3}}}\\&=&{\frac {3z^{2}-6z+3}{(1-z)^{6}}}\\&=&{\frac {3(z^{2}-2z+1)}{(1-z)^{6}}}\\&=&{\frac {3(1-z)^{2}}{(1-z)^{6}}}\\&=&{\frac {3}{(1-z)^{4}}}\\\end{matrix}}$

Differentiation technique

1.

${\begin{matrix}f(z)&=&{\frac {1}{(1-z)^{2}}}\\f'(z)&=&-{\frac {((1-z)^{2})'}{((1-z)^{2})^{2}}}\\&=&-{\frac {-2(1-z)}{(1-z)^{4}}}\\&=&{\frac {2}{(1-z)^{3}}}\end{matrix}}$

We use the result of the differentation of f(z)=(1-z)^n (f'(z) = -n(1-z)^n-1)

2.

${\begin{matrix}f(z)&=&{\frac {1}{(1-z)^{3}}}\\f'(z)&=&-{\frac {((1-z)^{3})\prime }{((1-z)^{3})^{2}}}\\&=&-{\frac {-3(1-z)^{2}}{(1-z)^{6}}}\\&=&{\frac {3}{(1-z)^{4}}}\end{matrix}}$

3.

${\begin{matrix}f(z)&=&{\frac {1}{(1+z)^{3}}}\\f'(z)&=&-{\frac {((1+z)^{3})'}{((1+z)^{3})^{2}}}\\&=&-{\frac {3(1+z)^{2}}{(1+z)^{6}}}\\&=&{\frac {-3}{(1+z)^{4}}}\end{matrix}}$

We use the result of exercise 3 of the previous section f(z)= (1+z)³ -> f'(z)=3(1+z)^2

4.

${\begin{matrix}f(z)&=&{\frac {1}{(1-z)^{n}}}\\f'(z)&=&-{\frac {((1-z)^{n})'}{((1-z)^{n})^{2}}}\\&=&-{\frac {-n(1-z)^{n-1}}{(1-z)^{2n}}}\\&=&-{\frac {-n}{(1-z)^{2n-(n-1)}}}\\&=&-{\frac {-n}{(1-z)^{2n-n+1}}}\\&=&{\frac {n}{(1-z)^{n+1}}}\end{matrix}}$

We use the result of the differentation of f(z)=(1-z)^n (f'(z) = -n(1-z)^n-1)