High School Mathematics Extensions/Supplementary/Differentiation/Solutions

Differentiate from first principle

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

2.

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

$f^{\prime }(z)=2z-2$

3.

$\begin{array}{ccc}{f}^{\prime }(z)& =& \underset{h\to 0}{\lim }\frac{\frac{1}{(1-z-h{)}^2}-\frac{1}{(1-z{)}^2}}{h}\\ & =& \underset{h\to 0}{\lim }\frac{1}{h}(\frac{1}{(1-z-h{)}^2}-\frac{1}{(1-z{)}^2})\\ & =& \underset{h\to 0}{\lim }\frac{1}{h}(\frac{(1-z{)}^2}{(1-z-h{)}^2(1-z{)}^2}-\frac{(1-z-h{)}^2}{(1-z-h{)}^2(1-z{)}^2})\\ & =& \underset{h\to 0}{\lim }\frac{1}{h}\frac{(1-z{)}^2-(1-z-h{)}^2}{(1-z-h{)}^2(1-z{)}^2}\\ & =& \underset{h\to 0}{\lim }\frac{1}{h}\frac{z^2-2z+1-(z^2+2hz-2z+h^2-2h+1)}{(1-z-h{)}^2(1-z{)}^2}\\ & =& \underset{h\to 0}{\lim }\frac{1}{h}\frac{z^2-2z+1-z^2-2hz+2z-h^2+2h-1)}{(1-z-h{)}^2(1-z{)}^2}\\ & =& \underset{h\to 0}{\lim }\frac{1}{h}\frac{-2hz-h^2+2h}{(1-z-h{)}^2(1-z{)}^2}\\ & =& \underset{h\to 0}{\lim }\frac{-2z-h+2}{(1-z-h{)}^2(1-z{)}^2}\\ & =& \frac{-2z+2}{(1-z{)}^2(1-z{)}^2}\\ & =& \frac{-2z+2}{(1-z{)}^4}\\ & =& \frac{2(1-z)}{(1-z{)}^4}\\ & =& \frac{2}{(1-z{)}^3}\end{array}$

4.

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

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

5. if

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

then

$\begin{array}{ccc}{f}^{\prime }(x)& =& \underset{k\to 0}{\lim }\frac{f(x+k)-f(x)}{k}\\ & =& \underset{k\to 0}{\lim }\frac{(g(x+k)+h(x+k))-(g(x)+h(x))}{k}\\ & =& \underset{k\to 0}{\lim }\frac{g(x+k)-g(x)+h(x+k)-h(x)}{k}\\ & =& \underset{k\to 0}{\lim }(\frac{g(x+k)-g(x)}{h}+\frac{h(x+k)-h(x)}{k})\\ & =& \underset{k\to 0}{\lim }\frac{g(x+k)-g(x)}{h}+\underset{k\to 0}{\lim }\frac{h(x+k)-h(x)}{k}\\ & =& g^{\prime }(x)+h^{\prime }(x)\end{array}$

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

1.

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

$\begin{array}{ccc}{f}^{\prime }(z)& =& -3z^2+6z-3\\ & =& -3(z^2-2z+1)\\ & =& -3(z-1{)}^2\end{array}$

2.

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

$\begin{array}{ccc}{f}^{\prime }(z)& =& 2z+2\\ & =& 2(z+1)\end{array}$

3.

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

$\begin{array}{ccc}{f}^{\prime }(z)& =& 3z^2+6z+3\\ & =& 3(z^2+2z+1)\\ & =& 3(z+1{)}^2\end{array}$

4.

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

$\begin{array}{ccc}{f}^{\prime }(z)& =& \underset{k\to 0}{\lim }\frac{\frac{1}{(1-z-k{)}^3}-\frac{1}{(1-z{)}^3}}{k}\\ & =& \underset{k\to 0}{\lim }\frac{1}{k}(\frac{1}{(1-z-k{)}^3}-\frac{1}{(1-z{)}^3})\\ & =& \underset{k\to 0}{\lim }\frac{1}{k}\frac{(1-z{)}^3-(1-z-k{)}^3}{(1-z-k{)}^3(1-z{)}^3}\\ & =& \underset{k\to 0}{\lim }\frac{1}{k}\frac{-z^3+3z^2-3z+1-(-z^3-3k{z}^2+3z^2-3k^{2}z+6kz-3z-k^3+3k^2-3k+1)}{(1-z-k{)}^3(1-z{)}^3}\\ & =& \underset{k\to 0}{\lim }\frac{1}{k}\frac{-z^3+3z^2-3z+1+z^3+3k{z}^2-3z^2+3k^{2}z-6kz+3z+k^3-3k^2+3k-1}{(1-z-k{)}^3(1-z{)}^3}\\ & =& \underset{k\to 0}{\lim }\frac{1}{k}\frac{3k{z}^2+3k^{2}z-6kz+k^3-3k^2+3k)}{(1-z-k{)}^3(1-z{)}^3}\\ & =& \underset{k\to 0}{\lim }\frac{3z^2+3kz-6z+k^2-3k+3)}{(1-z-k{)}^3(1-z{)}^3}\\ & =& \frac{3z^2-6z+3}{(1-z{)}^3(1-z{)}^3}\\ & =& \frac{3z^2-6z+3}{(1-z{)}^6}\\ & =& \frac{3(z^2-2z+1)}{(1-z{)}^6}\\ & =& \frac{3(1-z{)}^2}{(1-z{)}^6}\\ & =& \frac{3}{(1-z{)}^4}\end{array}$

Differentiation technique

1.

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

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

2.

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

3.

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

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

4.

$\begin{array}{ccc}f(z)& =& \frac{1}{(1-z{)}^n}\\ f^{\prime }(z)& =& -\frac{((1-z{)}^n{)}^{\prime }}{((1-z{)}^n{)}^2}\\ & =& -\frac{-n(1-z{)}^{n-1}}{(1-z{)}^{2n}}\\ & =& -\frac{-n}{(1-z{)}^{2n-(n-1)}}\\ & =& -\frac{-n}{(1-z{)}^{2n-n+1}}\\ & =& \frac{n}{(1-z{)}^{n+1}}\end{array}$

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