smc-1037-10-Sin/Cos Differentiation

Calculus, EXT1 C2 2025 HSC 14d

The function \(f(x)\) is defined by \(f(x)=\cos ^{-1}(\sin x)\) in the domain \((0, \pi)\).

Find \(f^{\prime}(x)\) for those values of \(x\) where it is defined. (3 marks)

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\(f^{\prime}(x)=-1 \ \text{for} \ \ x \in\left(0, \dfrac{\pi}{2}\right)\)

\(f^{\prime}(x)=1 \ \text{for} \ \ x \in\left(\dfrac{\pi}{2}, \pi\right)\)

Show Worked Solution

\(f(x)\)	\(=\cos ^{-1}(\sin x)\)
\(f^{\prime}(x)\)	\(=-\dfrac{\cos x}{\sqrt{1-\sin ^2 x}}\)
	\(=-\dfrac{\cos x}{\abs{\cos x}}\)
	\(= \pm 1\)

\(\text{Consider limitations:}\)

\(1-\sin ^2 x \neq 0 \ \Rightarrow \ \sin x \neq \pm 1 \ \Rightarrow \ x \neq \dfrac{\pi}{2}\)

\(\text{In lst quadrant:} \ \ -\dfrac{\cos x}{\abs{\cos x}}=-1\)

\(\text{In 2nd quadrant:}\ \ -\dfrac{\cos x}{\abs{\cos x}}=1\)

\(f^{\prime}(x)=-1 \ \ \text{for} \ \ x \in\left(0, \dfrac{\pi}{2}\right)\)

\(f^{\prime}(x)=1 \ \ \text{for} \ \ x \in\left(\dfrac{\pi}{2}, \pi\right)\)

Calculus, EXT1 C2 EQ-Bank 1

Given \(y=\dfrac{\cos 2 x}{x^2}\), find the gradient of the tangent of its inverse function at \(\left(\dfrac{8 \sqrt{2}}{\pi^2}, \dfrac{\pi}{4}\right)\). (3 marks)

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\(-\dfrac{\pi^2}{32}\)

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\(y=\dfrac{\cos2x}{x^2}\)

\(\text{Inverse: Swap} \ \ x \leftrightarrow y\)

\(x=\dfrac{\cos 2 y}{y^2}\)

\(u=\cos 2 y \ \ \quad \quad \quad v=y^2\)

\(v^{\prime}=-2 \sin 2 y \ \ \quad v^{\prime}=2 y\)

\(\dfrac{dx}{dy}=\dfrac{-2 y^2 \sin 2 y-2 y\, \cos 2 y}{y^4}\)

\(\dfrac{dy}{dx}=\dfrac{y^4}{-2 y^2 \sin 2 y-2 y\, \cos 2 y}\)

\(\text{At}\ \ y=\dfrac{\pi}{4}:\)

\(\dfrac{d y}{d x}\)	\(=\dfrac{\left(\dfrac{\pi}{4}\right)^4}{-2\left(\dfrac{\pi}{4}\right)^2 \sin \left(\dfrac{\pi}{2}\right)-2\left(\dfrac{\pi}{4}\right) \cos \left(\dfrac{\pi}{2}\right)}\)
	\(=-\dfrac{\pi^4}{256} \times \dfrac{8}{\pi^2}\)
	\(=-\dfrac{\pi^2}{32}\)

Calculus, EXT1 C2 2024 HSC 11e

Differentiate the function \(f(x)=\arcsin \left(x^5\right)\). (1 mark)

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\(f^{\prime}(x)=\dfrac{5 x^4}{\sqrt{1-x^{10}}}\)

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\(f(x)=\sin ^{-1}\left(x^5\right)\)

\(f^{\prime}(x)=5 x^4 \times \dfrac{1}{\sqrt{1-\left(x^5\right)^2}}=\dfrac{5 x^4}{\sqrt{1-x^{10}}}\)

Calculus, EXT1 C2 2020 HSC 13c

Suppose `f(x) = tan(cos^(-1)(x))` and `g(x) = (sqrt(1-x^2))/x`.

The graph of `y = g(x)` is given.

Show that `f^(′)(x) = g^(′)(x)`. (4 marks)

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Using part (i), or otherwise, show that `f(x) = g(x)`. (3 marks)

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`text(See Worked Solutions)`
`text(See Worked Solutions)`

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i. `f(x) = tan(cos^(-1)(x))`

♦ Mean mark (i) 50%.

`f^(′)(x)`	`= -1/sqrt(1-x^2) · sec^2(cos^(-1)(x))`
	`= -1/sqrt(1-x^2) · 1/(cos^2(cos^(-1)(x)))`
	`= -1/(x^2sqrt(1-x^2))`

`g(x) = (1-x^2)^(1/2) · x^(-1)`

`g^(′)(x)`	`= 1/2 · -2x(1-x^2)^(-1/2) · x^(-1)-(1-x^2)^(1/2) · x^(-2)`
	`= (-x)/(x sqrt(1-x^2))-sqrt(1-x^2)/(x^2)`
	`= (-x^2-sqrt(1-x^2) sqrt(1-x^2))/(x^2 sqrt(1-x^2))`
	`= (-x^2-(1-x^2))/(x^2sqrt(1-x^2))`
	`= -1/(x^2sqrt(1-x^2))`
	`=f^(′)(x)`

ii. `f^(′)(x) = g^(′)(x)`

♦♦♦ Mean mark (ii) 15%.

`=> f(x) = g(x) + c`

`text(Find)\ c:`

`f(1)`	`= tan(cos^(-1) 1)`
	`= tan 0`
	`= 0`

`g(1) = sqrt(1-1)/0 = 0`

`f(1) = g(1) + c`

`:. c = 0`

`:. f(x) = g(x)`

Calculus, EXT1 C2 2018 HSC 12c

Let `f(x) = sin^(-1) x + cos^(-1) x`.

Show that `f^{′}(x) = 0` (1 mark)

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Hence, or otherwise, prove

`qquad sin^(-1) x + cos^(-1) x = pi/2`. (1 mark)

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Hence, sketch

`qquad f(x) = sin^(-1) x + cos^(-1) x`. (1 mark)

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`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`text(See Worked Solutions)`

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i. `f^{′}(x) = 1/sqrt(1-x^2) + (-1/sqrt (1-x^2)) = 0`

ii. `text(S)text(ince)\ \ f^{′}(x) = 0\ \ => f(x)\ \ text(is a constant.)`

♦ Mean mark (ii) 37%.

`text(Substituting)\ \ x=1\ \ text{into the equation (any value works)}`

	`sin^(-1) 1 + cos^(-1) 1`	`= pi/2 + 0`
		`= pi/2\ \ text(… as required)`

iii. `text(Domain restrictions require:)\ \ -1<x<1`

♦ Mean mark (iii) 40%.

Calculus, EXT1 C2 2004 HSC 2b

Find `d/(dx)\ cos^(−1)\ (3x^2).` (2 marks)

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`(−6x)/(sqrt(1 − 9x^4))`

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`d/(dx)\ cos^(−1)\ (3x^2)`

`= (−1)/sqrt(1 − (3x^2)^2) xx d/(dx) (3x^2)`

`= (−6x)/(sqrt(1 − 9x^4))`

Calculus, EXT1 C2 2006 HSC 2a

Let `f(x) = sin^-1 (x + 5).`

State the domain and range of the function `f(x).` (2 marks)

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Find the gradient of the graph of `y = f(x)` at the point where `x = -5.` (2 marks)

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Sketch the graph of `y = f(x).` (2 marks)

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`text(Domain):\ -6 <= x <= -4, \ \ \ text(Range):\ -pi/2 <= y <= pi/2`
`1`

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i. `f(x) = sin^-1 (x + 5)`

`text(Domain)`

`-1 <= x + 5 <= 1`

`-6 <= x <= -4`

`text(Range)`

`-pi/2 <= y <= pi/2`

ii. `y = sin^-1 (x + 5)`

`(dy)/(dx) = 1/sqrt(1-(x + 5)^2)`

`text(When)\ \ x = -5`

`(dy)/(dx)`	`= 1/sqrt(1-(-5 + 5)^2)`
	`= 1/sqrt(1-0)`
	`= 1`

`:.\ text(Gradient of)\ \ y = f(x)\ \ text(at)\ \ x = -5\ \ text(is)\ \ 1.`

iii.

Calculus, EXT1 C2 2005 HSC 2a

Find `d/(dx) (2 sin^-1 5x).` (2 marks)

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`10/(sqrt (1 – 25x^2))`

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`d/(dx) (2 sin^-1 5x)`	`= 2 xx d/(dx) (sin^-1 5x)`
	`= 2 xx 1/(sqrt(1 – (5x)^2)) xx d/(dx) (5x)`
	`= 2/(sqrt (1 – 25x^2)) xx 5`
	`= 10/(sqrt(1 – 25 x^2))`

Calculus, EXT1 C2 2015 HSC 13d

Let `f(x) = cos^(-1)\ (x) + cos^(-1)\ (-x)`, where `-1 ≤ x ≤ 1`.

By considering the derivative of `f(x)`, prove that `f(x)` is constant. (2 marks)

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Hence deduce that `cos^(-1)\ (-x) = pi - cos^(-1)\ (x)`. (1 mark)

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`text(See Worked Solutions.)`
`text(See Worked Solutions.)`

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i. `text(Prove)\ f(x)\ text(is a constant)`

`f(x)`	`= cos^(-1)(x) + cos^(-1)(-x), \ \ -1 ≤ x ≤ 1`
`f′(x)`	`= (-1)/sqrt(1 – x^2) + (-1)/sqrt(1 -(-x)^2) xx d/(dx) (-x)`
	`= (-1)/sqrt(1 – x^2) + 1/sqrt(1 – x^2)`
	`= 0`

`:.\ text(S)text(ince)\ \ f′(x) = 0, f(x)\ text(must be a constant.)`

♦ Mean mark 35%.

ii.	`f(0)`	`= cos^(−1)(0) + cos^(−1)(0)`
		`= pi/2 + pi/2`
		`= pi`

`:.f(x) = pi`

`pi = cos^(−1)(x) + cos^(−1)(−x)`

`:.cos^(−1)(−x) = pi – cos^(−1)(x)\ \ …text(as required)`

Calculus, EXT1 C2 2008 HSC 1b

Differentiate `cos^(–1) (3x)` with respect to `x`. (2 marks)

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`(-3)/sqrt(1 – 9x^2)`

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`y`	`= cos^(-1) (3x)`
`dy/dx`	`= – 1/sqrt(1 – (3x)^2) * d/(dx) (3x)`
	`= (-3)/sqrt(1 – 9x^2)`

Calculus, EXT1 C2 2014 HSC 6 MC

What is the derivative of `3 sin^(-1)\ x/2`?

`6/sqrt(4 - x^2)`
`3/sqrt(4 - x^2)`
`3/(2sqrt(4 - x^2))`
`3/(4sqrt(4 - x^2))`

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`B`

Show Worked Solution

`y = 3 sin^(-1)\ x/2`

`dy/dx = 3 xx 1/sqrt(4 – x^2)`

`=> B`

Calculus, EXT1 C2 2013 HSC 11g

Differentiate `x^2 sin^(–1) 5x`. (2 marks)

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`(5x^2)/sqrt(1\ – 25x^2) + 2x sin^(-1) 5x`

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`y`	`= x^2 sin^(-1) 5x`
`text(Using the product rule)`
`(dy)/(dx)`	`= x^2 xx 1/sqrt(1\ – (5x)^2) xx d/(dx)(5x) + 2x sin^(-1) 5x`
	`= (5x^2)/sqrt(1\ – 25x^2) + 2x sin^(-1) 5x`

Calculus, EXT1 C2 2012 HSC 9 MC

What is the derivative of `cos^(–1) (3x)`?

`1/(3 sqrt(1 - 9x^2))`
`(-1)/(3 sqrt(1 - 9x^2))`
`3/sqrt(1 - 9x^2)`
`(-3)/sqrt(1 - 9x^2)`

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`D`

Show Worked Solution

`y = cos^(-1) (3x)`

`dy/dx`	`= (-1)/sqrt(1\ – (3x)^2) xx d/dx (3x)`
	`= (-3)/sqrt(1\ – 9x^2)`

`=> D`