Standard Differentiation

Calculus, 2ADV C1 EQ-Bank 13

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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a. `y^{′} = 4x+5`

b. `y = 9x-2`

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a.	`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`

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

`y^{′} = 4+5 = 9`

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

Calculus, 2ADV C1 EQ-Bank 21

Use differentiation by first principles to find \(y^{\prime}\) given \(y=\dfrac{5}{x}\). (3 marks)

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\(\text{See worked solutions}\)

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\(\dfrac{d}{d x}\left(\dfrac{5}{x}\right)\)	\(=\displaystyle \lim _{h \rightarrow 0} \dfrac{f(x+h)-f(x)}{h}\)
	\(=\displaystyle \lim _{h \rightarrow 0}\dfrac{\dfrac{5}{x+h}-\dfrac{5}{x}}{h}\)
	\(=\displaystyle \lim _{h \rightarrow 0} \dfrac{\dfrac{5 x-5(x+h)}{x(x+h)}}{h}\)
	\(=\displaystyle \lim _{h \rightarrow 0} \dfrac{\dfrac{-5 h}{x(x+h)}}{h}\)
	\(=\displaystyle \lim _{h \rightarrow 0} \dfrac{-5}{x(x+h)}\)
	\(=\dfrac{-5}{x^2}\)

Calculus, 2ADV C1 EQ-Bank 20

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`

Show Worked Solution

`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 4 MC

The derivative of \(n x^{2n+1}\) can be expressed as

\(2 n^2 x^{2 n+1}\)
\(2 n^2 x^{2 n}\)
\((2 n+1) n x^{2 n}\)
\((2 n+1) n x^{2 n+1}\)

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\(C\)

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\(y\)	\(=n x^{2n+1}\)
\(y^{′}\)	\(=(2 n+1) n x^{2n+1-1}\)
	\(=(2 n+1) n x^{2 n}\)

\(\Rightarrow C\)

Calculus, 2ADV C1 2023 MET2 11 MC

Two functions, \(f\) and \(g\), are continuous and differentiable for all \(x\in R\). It is given that \(f(-2)=-7,\ g(-2)=8\) and \(f^{′}(-2)=3,\ g^{′}(-2)=2\).

The gradient of the graph \(y=f(x)\times g(x)\) at the point where \(x=-2\) is

\(-6\)
\(0\)
\(6\)
\(10\)

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\(D\)

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\(\text{Using the Product Rule when}\ \ x=-2:\)

\(\dfrac{d}{dx}(f(x)\times g(x))\)	\(=f(x)g^{′}(x)+g(x)f^{′}(x)\)
	\(=f(-2)g^{′}(-2)+g(-2)f^{′}(-2)\)
	\(=-7\times 2+8\times 3\)
	\(=10\)

\(\Rightarrow D\)

♦♦♦ Mean mark 22%.

Calculus, 2ADV C1 2023 HSC 7 MC

It is given that `y=f(g(x))`, where `f(1)=3`, `f^{′}(1)=-4`, `g(5)=1` and `g^{′}(5)=2`.

What is the value of `y^{′}` at `x=5`?

`-8`
`-4`
`3`
`6`

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`A`

Show Worked Solution

`y`	`=f(g(x))`
`y^{′}`	`=f^{′}(g(5)) xx g^{′}(5)`
	`=f^{′}(1) xx 2`
	`=-4xx2`
	`=-8`

`=>A`

♦♦ Mean mark 38%.

Calculus, 2ADV C1 EQ-Bank 24

Evaluate `f^{′}(4)`, where `f(x) = xsqrt(2x + 1)`. (3 marks)

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`13/3`

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`f(x) = xsqrt(2x + 1)`

`f^{′}(x)`	`=1 sqrt(2x + 1) + x xx 1/2 xx 2(2x+1)^(-1/2)`
	`=sqrt(2x + 1) + x(2x+1)^(-1/2)`
`f^{′}(4)`	`= sqrt9 + 4(9)^(-1/2)`
	`=3 + 4/3`
	`=13/3`

Calculus, 2ADV C1 EQ-Bank 12

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 2019 HSC 11c

Differentiate `(2x + 1)/(x + 5)`. (2 marks)

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`9/(x + 5)^2`

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`text(Using quotient rule:)`

`u=2x+1,`	`v=x+5`
`u^{′} = 2,`	`v^{′} = 1`

`y^{′}`	`= (u^{′} v-v^{′} u)/v^2`
	`= (2(x + 5)-(2x + 1))/(x + 5)^2`
	`= (2x + 10-2x-1)/(x + 5)^2`
	`= 9/(x + 5)^2`

Calculus, 2ADV C1 EQ-Bank 26

Let `g(x) = (2-x^3)^3`.

Evaluate `g^{′}(-1)`. (2 marks)

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`-81`

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`text(Using Chain Rule:)`

`g^{′}(x)`	`= 3 (2-x^3)^2 (-3x^2)`
	`= -9x^2 (2-x^3)^2`
`:. g^{′}(1)`	`= -9 (-1)^2 [2-(-1)^3]^2`
	`= -81`

Calculus, 2ADV C1 2016 HSC 11b

Differentiate `(x + 2)/(3x-4).` (2 marks)

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`(-10)/(3x-4)^2`

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`y = (x + 2)/(3x-4)`

`text(Using the quotient rule:)`

`(g/h)^{′}`	`= (g^{′} h-g h^{′})/h^2`
`y prime`	`= (1 (3x-4)-(x + 2) · 3)/(3x-4)^2`
	`= (-10)/(3x-4)^2`

Calculus, 2ADV C1 2015 HSC 12c

Find `f^{′}(x)`, where `f(x) = (x^2 + 3)/(x-1).` (2 marks)

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`((x-3) (x + 1))/(x-1)^2`

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`f(x) = (x^2 + 3)/(x-1)`

`text(Using the quotient rule:)`

`u`	`= x^2 + 3`	`\ \ \ \ \ \ v`	`= x-1`
`u^{′}`	`= 2x`	`\ \ \ \ \ \ v^{′}`	`= 1`

`f^{′}(x)`	`= (u^{′} v-uv^{′})/v^2`
	`= (2x (x-1)-(x^2 + 3) xx 1)/(x-1)^2`
	`= (2x^2-2x-x^2-3)/(x-1)^2`
	`= (x^2-2x-3)/(x-1)^2`
	`= ((x-3) (x + 1))/(x-1)^2`

Calculus, 2ADV C1 2005 HSC 2bii

Differentiate with respect to `x`:

`x^2/(x-1).` (2 marks)

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`(x(x-2))/(x-1)^2`

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`y = x^2/(x-1)`

`text(Using)\ \ dy/dx = (u^{′}v-uv^{′})/v^2:`

`u`	`= x^2`	`v`	`= x-1`
`u^{′}`	`= 2x`	`v^{′}`	`= 1`

`dy/dx`	`= (2x(x-1)-x^2(1))/(x-1)^2`
	`= (2x^2-2x-x^2)/(x-1)^2`
	`= (x^2-2x)/(x-1)^2`
	`= (x(x-2))/(x-1)^2`

Calculus, 2ADV C1 2004 HSC 1b

Differentiate `x^4 + 5x^(−1)` with respect to `x`. (2 marks)

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`4x^3 – 5x^(-2)`

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`y`	`= x^4 + 5x^(-1)`
`dy/dx`	`= 4x^3 – 5x^(-2)`

Calculus, 2ADV C1 2014 HSC 11c

Differentiate `x^3/(x + 1)`. (2 marks)

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`(x^2(2x + 3))/((x + 1)^2)`

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`y = (x^3)/(x + 1)`

`text(Using)\ \ \ dy/dx = (u^{′} v\ – uv^{′})/(v^2)`

`u`	`=x^3`	`\ \ \ \ \ v`	`=(x+1)`
`u^{′}`	`=3x^2`	`v^{′}`	`=1`

`dy/dx`	`= (3x^2 (x + 1)-x^3 (1))/((x + 1)^2)`
	`= (3x^3 + 3x^2-x^3)/((x + 1)^2)`
	`= (2x^3 + 3x^2)/((x + 1)^2)`
	`= (x^2 (2x + 3))/((x + 1)^2)`

Calculus, 2ADV C3 2010 HSC 8d

Let `f(x) = x^3-3x^2 + kx + 8`, where `k` is a constant.

Find the values of `k` for which `f(x)` is an increasing function. (2 marks)

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`k>3`

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`f(x)`	`= x^3-3x^2 + kx + 8`
`f^{′}(x)`	`= 3x^2-6x + k`

`f(x)\ text(is increasing when)\ \ f^{′}(x) > 0`

`=> 3x^2-6x + k > 0`

♦♦ Mean mark 28%.
MARKER’S COMMENT: The arithmetic required to solve `36-12k<0` proved the undoing of many students.

`f^{′}(x)\ text(is always positive)`

`=> f^{′}(x)\ text(is a positive definite.)`

`text(i.e. when)\ \ a > 0\ text(and)\ Delta < 0`

`a=3>0`

`Delta = b^2-4ac`

`(-6)^2-(4 xx 3 xx k)`	`<0`
`36-12k`	`<0`
`12k`	`>36`
`k`	`>3`

`:.\ f(x)\ text(is increasing when)\ \ k > 3.`

Calculus, 2ADV C1 2013 HSC 11b

Evaluate `lim_(x->2) ((x-2)(x+2)^2)/(x^2-4)`. (2 marks)

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`4`

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`lim_(x ->2) ((x-2)(x+2)^2)/(x^2-4)`

COMMENT: This question has been simplified as students no longer need to factorise the difference between 2 cubes (`x^3-2^3`).

`=lim_(x->2) ( (x -2)(x+2)^2)/( (x-2)(x+2)`

`=lim_(x->2) (x+2)`

`=4`