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Answer :
To find the derivative of a function using first principles, we use the definition of the derivative:
[tex]f'(x) = \lim_{{h \to 0}} \frac{f(x+h) - f(x)}{h}[/tex]
Let's apply this to each function given:
a. [tex]f(x) = 2x[/tex]
Compute [tex]f(x+h)[/tex]:
[tex]f(x+h) = 2(x+h) = 2x + 2h[/tex]Substitute into the derivative formula:
[tex]f'(x) = \lim_{{h \to 0}} \frac{2x + 2h - 2x}{h} = \lim_{{h \to 0}} \frac{2h}{h} = \lim_{{h \to 0}} 2 = 2[/tex]
So, [tex]f'(x) = 2[/tex].
b. [tex]f(x) = 3 - x^2[/tex]
Compute [tex]f(x+h)[/tex]:
[tex]f(x+h) = 3 - (x+h)^2 = 3 - (x^2 + 2xh + h^2) = 3 - x^2 - 2xh - h^2[/tex]Substitute into the derivative formula:
[tex]f'(x) = \lim_{{h \to 0}} \frac{3 - x^2 - 2xh - h^2 - (3 - x^2)}{h} = \lim_{{h \to 0}} \frac{-2xh - h^2}{h}[/tex]
[tex]= \lim_{{h \to 0}} (-2x - h) = -2x[/tex]
So, [tex]f'(x) = -2x[/tex].
c. [tex]f(x) = \frac{2}{x}[/tex]
Compute [tex]f(x+h)[/tex]:
[tex]f(x+h) = \frac{2}{x+h}[/tex]Substitute into the derivative formula:
[tex]f'(x) = \lim_{{h \to 0}} \frac{\frac{2}{x+h} - \frac{2}{x}}{h} = \lim_{{h \to 0}} \frac{2x - 2(x+h)}{h(x+h)x}[/tex]
[tex]= \lim_{{h \to 0}} \frac{-2h}{h(x^2 + xh)} = \lim_{{h \to 0}} \frac{-2}{x^2 + xh} = -\frac{2}{x^2}[/tex]
So, [tex]f'(x) = -\frac{2}{x^2}[/tex].
d. [tex]g(x) = -\frac{3}{x}[/tex]
Compute [tex]g(x+h)[/tex]:
[tex]g(x+h) = -\frac{3}{x+h}[/tex]Substitute into the derivative formula:
[tex]g'(x) = \lim_{{h \to 0}} \frac{-\frac{3}{x+h} + \frac{3}{x}}{h} = \lim_{{h \to 0}} \frac{-3x + 3(x+h)}{h(x+h)x}[/tex]
[tex]= \lim_{{h \to 0}} \frac{3h}{h(x+h)x} = \lim_{{h \to 0}} \frac{3}{-x^2 - xh} = \frac{3}{x^2}[/tex]
So, [tex]g'(x) = \frac{3}{x^2}[/tex].
These calculations show how the derivative is found using first principles, breaking down each function to find its rate of change at any point.
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