High School Mathematics Extensions/Supplementary/Differentiation/Solutions

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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)