Answer :
Let's address each part of the question one by one:
10.1.1 Determine [tex]f'(x)[/tex] from first principles
The function given is [tex]f(x) = -\frac{2}{x}[/tex].
To find the derivative [tex]f'(x)[/tex] from first principles, use the definition of the derivative:
[tex]f'(x) = \lim_{{h \to 0}} \frac{f(x+h) - f(x)}{h}[/tex]
Substituting [tex]f(x) = -\frac{2}{x}[/tex]:
[tex]f(x+h) = -\frac{2}{x+h}[/tex]
Thus,
[tex]f'(x) = \lim_{{h \to 0}} \frac{-\frac{2}{x+h} + \frac{2}{x}}{h}[/tex]
Combine the fractions:
[tex]f'(x) = \lim_{{h \to 0}} \frac{2x - 2(x+h)}{h(x+h)x}[/tex]
Simplify the numerator:
[tex]= \lim_{{h \to 0}} \frac{2x - 2x - 2h}{h(x+h)x} = \lim_{{h \to 0}} \frac{-2h}{h(x+h)x}[/tex]
Cancel [tex]h[/tex] in the numerator and the denominator:
[tex]= \lim_{{h \to 0}} \frac{-2}{(x+h)x}[/tex]
Taking the limit as [tex]h \to 0[/tex], we get:
[tex]f'(x) = -\frac{2}{x^2}[/tex]
10.1.2 For which value(s) of [tex]x[/tex] will [tex]f'(x) > 0[/tex]? Justify your answer.
We have found that [tex]f'(x) = -\frac{2}{x^2}[/tex].
Since [tex]x^2[/tex] is always positive for any real number except zero, [tex]f'(x) = -\frac{2}{x^2}[/tex] will always be negative for any real number except zero, meaning there are no values of [tex]x[/tex] for which [tex]f'(x) > 0[/tex].
10.2 Evaluate [tex]\frac{dy}{dx}[/tex] if [tex]y = \frac{1}{4}x^2 - 2x[/tex].
To find the derivative [tex]\frac{dy}{dx}[/tex], differentiate each term with respect to [tex]x[/tex]:
- The derivative of [tex]\frac{1}{4}x^2[/tex] is [tex]\frac{1}{4} \times 2x = \frac{1}{2}x[/tex].
- The derivative of [tex]-2x[/tex] is [tex]-2[/tex].
Thus, [tex]\frac{dy}{dx} = \frac{1}{2}x - 2[/tex].
10.3 Given: [tex]y = 4\sqrt{x^2}[/tex] and [tex]x = w^{-3}[/tex] Determine [tex]\frac{dy}{dw}[/tex].
Start by simplifying [tex]y = 4\sqrt{x^2}[/tex]:
[tex]\sqrt{x^2} = |x|[/tex], so [tex]y = 4|x|[/tex].
Substitute [tex]x = w^{-3}[/tex]:
[tex]y = 4|w^{-3}|[/tex], which equals [tex]4w^{-3}[/tex] since [tex]w^{-3}[/tex] is non-negative.
Now differentiate with respect to [tex]w[/tex]:
The derivative [tex]\frac{d}{dw}(4w^{-3})[/tex] is:
[tex]4 \times (-3)w^{-4} = -12w^{-4}[/tex].
Therefore, [tex]\frac{dy}{dw} = -12w^{-4}[/tex].
