MATH110 December 2013
• Q1 (a) • Q1 (b) • Q1 (c) • Q1 (d) • Q2 • Q3 (a) • Q3 (b) • Q4 (a) • Q4 (b) • Q4 (c) • Q5 (a) • Q5 (b) • Q6 (a) • Q6 (b) • Q6 (c) • Q7 • Q8 • Q9 • Q10 (a) • Q10 (b) • QS01 10(a) • QS01 10(b) •
Question 03 (b)

Rewrite $f(x)$ in such a way that you can differentiate it using a different rule than the Power Rule. Name the rule of differentiation which is applicable, and then confirm that the derivative calculated using this alternative method is the same as in part (a).
(Bonus) You will receive one bonus mark for each different differentiation rule you use to differentiate $\displaystyle f(x)$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint

There are lots of choices here. The easier ones to see are the chain rule, the definition of a derivative, and using logarithmic differentiation. There are of course many others as you will see in the solution.

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Solution

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Parts b and c combined: (These are all the rules I can think of).
Addition rule:
${\begin{aligned}f(x)&={\frac {1}{2}}x^{6}+{\frac {1}{2}}x^{6}&f'(x)&=3x^{5}+3x^{5}\end{aligned}}$
Subtraction rule:
${\begin{aligned}f(x)&=2x^{6}x^{6}&f'(x)&=12x^{5}6x^{5}\end{aligned}}$
Product rule:
${\begin{aligned}f(x)&=x^{2}\cdot x^{4}&f'(x)&=2x\cdot x^{4}+4x^{3}\cdot x^{2}\end{aligned}}$
Quotient rule:
${\begin{aligned}f(x)&={\frac {x^{7}}{x}}&f'(x)&={\frac {7x^{6}\cdot xx^{7}\cdot 1}{x^{2}}}\end{aligned}}$
Chain rule:
${\begin{aligned}f(x)&=(x^{3})^{2}&f'(x)&=3x^{2}\cdot 2(x^{3})\end{aligned}}$
Logdiff:
${\begin{aligned}\ln(f(x))&=6\ln(x)&f'(x)&={\frac {6}{x}}\cdot x^{6}\end{aligned}}$
Expchain rule:
${\begin{aligned}f(x)&=\exp[6\ln(x)]&f'(x)&={\frac {6}{x}}\cdot \exp[6\ln(x)]\end{aligned}}$
Limit definition of derivative:
${\begin{aligned}f'(x)&=\lim _{h\rightarrow 0}{\frac {(x+h)^{6}x^{6}}{h}}&f'(x)&=\lim _{h\rightarrow 0}{\frac {6x^{5}h+15x^{4}h^{2}+\ldots +h^{6}}{h}}\\f'(x)&=\lim _{a\rightarrow x}{\frac {a^{6}x^{6}}{ax}}&f'(x)&=\lim _{a\rightarrow x}{\frac {(ax)(a^{5}+a^{4}x+\ldots x^{5})}{(ax)}}\end{aligned}}$

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