How many QCAA Mathematical Methods questions cover Differentiation of trigonometric functions and differentiation rules?

Q: How many QCAA Mathematical Methods questions cover Differentiation of trigonometric functions and differentiation rules?

AusGrader has 56 QCAA Mathematical Methods questions on Differentiation of trigonometric functions and differentiation rules, all with instant AI grading and detailed marking feedback.

QCAA (QCE) Mathematical Methods — Differentiation of trigonometric functions and differentiation rules Questions & Answers | AusGrader

Q2

2024

QCAA

Paper 1

1 mark

Q2

1 mark

Determine $\frac{dy}{dx}$ for the function $y = e^{\sin(x)}$

A

$\cos(x) e^{\sin(x)}$

B

$\sin(x) e^{\cos(x)}$

C

$e^{\sin(x)}$

D

$e^{\cos(x)}$

Q6

2024

QCAA

Paper 1

1 mark

Q6

1 mark

Differentiate $y = \ln(x) \cos(x)$ with respect to $x$ .

A

$\frac{\cos(x)}{x}$

B

$-\frac{\sin(x)}{x}$

C

$\frac{\cos(x)}{x} + \ln(x) \sin(x)$

D

$\frac{\cos(x)}{x} - \ln(x) \sin(x)$

Q13

2024

QCAA

Paper 2

8 marks

Q13

The number of termites in a particular nest can be modelled by $N(t) = \frac{A}{2 + e^{-t}}$ , where $A$ is a constant and $t$ represents time (months) since the nest first became a visible mound above ground level.
It is estimated that when the mound first became visible, the population was $3 \times 10^5$ termites.

Q13a

1 mark

Determine the value of $A$ .

Descriptor	Marks
Correctly determines the value of $A$	1

Q13b

2 marks

Determine the number of termites in the nest half a year after the mound became visible.

$N(t)=\dfrac{9\times 10^5}{2+e^{-t}}$

$N(6)=\dfrac{9\times 10^5}{2+e^{-6}}$

$=449\ 442.9711$

$=449\ 443\ \text{termites}$

Marking Criteria

Descriptor	Marks
Correctly determines $t=6$	1
Determines the number of termites	1

Q13c

2 marks

Determine the time in months after the mound became visible for the initial population to increase by 130 000 termites. Express the time as a decimal.

$300\ 000+130\ 000=430\ 000$

$N(t)=\dfrac{9\times 10^5}{2+e^{-t}}$

$430\ 000=\dfrac{9\times 10^5}{2+e^{-t}}$

Using a GDC:

$t=2.3749\ \text{months}$

Marking Criteria

Descriptor	Marks
Correctly determines the required number of termites	1
Determines the time required	1

Q13d

2 marks

Develop a formula for the rate of change in the number of termites at any time after the mound became visible. Express your formula as a fraction.

$N(t)=\dfrac{9\times 10^5}{2+e^{-t}}$

$N(t)=(9\times 10^5)(2+e^{-t})^{-1}$

$\dfrac{dN}{dt}=-(9\times 10^5)(2+e^{-t})^{-2}\times -e^{-t}$

$\dfrac{dN}{dt}=\dfrac{(9\times 10^5)}{e^t(2+e^{-t})^2}$

Marking Criteria

Descriptor	Marks
Shows application of the chain rule	1
Determines a formula for the number of termites expressed as a fraction	1

Q13e

1 mark

Determine the rate of change in the number of termites five months after the mound became visible.

Using the formula developed in d) or a GDC:

$N'(5)=1505.8744$

$=1506\ \text{termites/month}$

Marking Criteria

Descriptor	Marks
Determines the rate of change	1

Q13

2022

QCAA

Paper 1

9 marks

Q13a

1 mark

Determine the derivative of $f(x) = 3e^{2x+1}$

Descriptor	Marks
correctly determines the derivative	1

Q13b

3 marks

Given that $g(x) = \frac{\ln(x)}{x}$ , determine the simplest value of $g'(e)$ .

$g(x) = \frac{\ln(x)}{x}$
Let $u = \ln(x)$ and $v = x$
$\therefore \frac{du}{dx} = \frac{1}{x}$ and $\frac{dv}{dx} = 1$
$g'(x) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$
$= \frac{x \times \frac{1}{x} - \ln(x) \times 1}{x^2}$
$= \frac{1-\ln(x)}{x^2}$

$g'(e) = \frac{1-\ln(e)}{(e)^2} = 0$

Marking Criteria

Descriptor	Marks
correctly identifies the use of the product or quotient rule	1
correctly determines the derivative	1
determines the derivative at the given value	1

Q13c

5 marks

Determine the second derivative of $h(x) = x\sin(x)$ . (Give your answer in simplest form.)

$h(x) = x \sin(x)$

Let $u = x$ and $v = \sin(x)$
$\therefore \frac{du}{dx} = 1$ and $\frac{dv}{dx} = \cos(x)$
$h'(x) = u\frac{dv}{dx} + v\frac{du}{dx}$
$= x \times \cos(x) + \sin(x) \times 1$
$= x \cos(x) + \sin(x) \times 1$

$= x \cos(x) + \sin(x)$
Let $u = x$ and $v = \cos(x)$
$\therefore \frac{du}{dx} = 1$ and $\frac{dv}{dx} = -\sin(x)$
$h''(x) = u\frac{dv}{dx} + v\frac{du}{dx} + \cos(x)$
$= x \times -\sin(x) + \cos(x) \times 1 + \cos(x)$

$= 2 \cos(x) - x \sin(x)$

Marking Criteria

Descriptor	Marks
correctly identifies the use of the product rule	1
correctly determines the derivative	1
correctly identifies the use of the product rule and that $\frac{d}{dx}(h(x) + g(x)) = \frac{d}{dx}h(x) + \frac{d}{dx}g(x)$	1
determines the second derivative	1
simplifies the second derivative	1

Q14

2021

QCAA

Paper 1

4 marks

Q14

Consider the function $f(x)=\ln(3x+4)$ , for $x > \frac{-4}{3}$

Q14a

1 mark

Determine $f'(x)$ .

Descriptor	Marks
correctly determines the derivative	1

Q14b

2 marks

Determine the $x$ -intercept of the graph of $f(x)$ .

$x$ -intercept $(y = 0)$
$0 = \ln(3x + 4)$
$3x + 4 = 1$
$x = -1$

Marking Criteria

Descriptor	Marks
correctly determines the linear equation	1
correctly determines the x-intercept	1

Q14c

1 mark

Determine the gradient of the tangent to the graph of $f(x)$ at the $x$ -intercept.

QCAA Mathematical Methods Differentiation of trigonometric functions and differentiation rules

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