High School Mathematics Extensions/Supplementary/Differentiation/Solutions

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^2z +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^2z -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^2z -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)