smc-1069-40-1st Principles

Calculus, 2ADV C1 EQ-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)=5x^2-2x`. (2 marks)

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`f(x)=5x^2-2x`

`f^{′}(x)`	`= \lim_{h->0} \frac{5(x+h)^2-2(x+h)-(5x^2-2x)}{h}`
	`= \lim_{h->0} \frac{5x^2+10xh+h^2-2x-2h-5x^2+2x}{h}`
	`= \lim_{h->0} \frac{10xh+h^2-2h}{h}`
	`= \lim_{h->0} \frac{h(10x+h-2)}{h}`
	`= \lim_{h->0} 10x+h-2`
	`=10x-2`

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`f(x)=5x^2-2x`

`f^{′}(x)`	`= \lim_{h->0} \frac{5(x+h)^2-2(x+h)-(5x^2-2x)}{h}`
	`= \lim_{h->0} \frac{5x^2+10xh+h^2-2x-2h-5x^2+2x}{h}`
	`= \lim_{h->0} \frac{10xh+h^2-2h}{h}`
	`= \lim_{h->0} \frac{h(10x+h-2)}{h}`
	`= \lim_{h->0} 10x+h-2`
	`=10x-2`

Calculus, 2ADV C1 EQ-Bank 2

When differentiating `f(x) = 3-2x-x^2` from first principles, a student began the solution as shown below.

Complete the solution. (2 marks)

`f^{′}(x) = lim_(h->0) (f(x + h)-f(x))/h`

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`f^{′}(x) = -2x-2`

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	`f^{′}(x)`	`= lim_(h->0) (f(x + h)-f(x))/h`
		`= lim_(h->0) (3-2(x + h) – (x+h)^2-(3-2x-x^2))/h`
		`= lim_(h->0) (3-2x-2h-x^2-2hx-h^2 – 3 + 2x + x^2)/h`
		`= lim_(h->0) (-2h-2hx-h^2)/h`
		`= lim_(h->0) (h(-2x-2-h))/h`

`:.\ f^{′}(x) = -2x-2`

Calculus, 2ADV C1 EQ-Bank 3

Use differentiation by first principles to find `y^{′}`, given `y = 2x^2 + 5x`. (2 marks)

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Find the equation of the tangent to the curve when `x = 1`. (1 mark)

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`y^{′} = 4x + 5`
`y = 9x-2`

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i.	`f(x)`	`= 2x^2 + 5x`
	`f^{′}(x)`	`= lim_(h->0) (f(x + h)-f(x))/h`
		`= lim_(h->0) ((2(x + h)^2 + 5(x + h))-(2x^2 + 5x))/h`
		`= lim_(h->0)(2x^2 + 4xh + 2h^2 + 5x + 5h-2x^2-5x)/h`
		`= lim_(h->0)(4xh + 2h^2 + 5h)/h`
		`= lim_(h->0)(h(4x + 5 + 2h))/h`

`:.\ y^{′} = 4x + 5`

ii. `text(When)\ \ x = 1, y = 7`

`y^{′} = 4 + 5 = 9`

`:. y-7`	`= 9(x-1)`
`y`	`= 9x-2`