Calculus, 2ADV C1 EO-Bank 6

Use the definition of the derivative, `f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}` to find `f^{\prime}(x)` if `f(x)=x-3x^2`. (2 marks)

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`f(x)=x-3x^2`

`f^{′}(x)`	`= \lim_{h->0} \frac{(x+h)-3(x+h)^2-(x-3x^2)}{h}`
	`= \lim_{h->0} \frac{x+h-3x^2-6hx-3h^2-x+3x^2}{h}`
	`= \lim_{h->0} \frac{h-6hx-3h^2}{h}`
	`= \lim_{h->0} \frac{h(1-6x-3h)}{h}`
	`= \lim_{h->0} 1-6x-3h`
	`=1-6x`

Show Worked Solution

`f(x)=x-3x^2`

`f^{′}(x)`	`= \lim_{h->0} \frac{(x+h)-3(x+h)^2-(x-3x^2)}{h}`
	`= \lim_{h->0} \frac{x+h-3x^2-6hx-3h^2-x+3x^2}{h}`
	`= \lim_{h->0} \frac{h-6hx-3h^2}{h}`
	`= \lim_{h->0} \frac{h(1-6x-3h)}{h}`
	`= \lim_{h->0} 1-6x-3h`
	`=1-6x`