Question 1
If $f(x)=\sin(3x)$, determine the value of $f'\left(\dfrac{\pi}{8}\right)$.
Not quite. 2.772 is $3\sin\left(\tfrac{3\pi}{8}\right)$. The chain rule’s 3 is there, but sine should have become cosine. It’s also $3\cos\left(\tfrac{\pi}{8}\right)$, the same number, which is what you get if the 3 inside the cosine is dropped. In 2023, 7.2% of Queensland students chose (A).
That’s it. Step through the hints below to see why. In 2023, 83.2% of Queensland students chose (B).
Not quite. 0.929 isn’t what $3\cos(3x)$ gives at $x=\tfrac{\pi}{8}$, in radians or in degrees. Check both steps: differentiate with the chain rule, then substitute with your calculator in radian mode. In 2023, 3.4% of Queensland students chose (C).
Not quite. 0.383 is $\cos\left(\tfrac{3\pi}{8}\right)$. The chain rule also multiplies by the derivative of the inside, $3x$, so $f'(x)=3\cos(3x)$. In 2023, 5.9% of Queensland students chose (D).
This is the question’s point. The line just touches the curve, so it’s the tangent. Its gradient is $3\cos\left(\tfrac{3\pi}{8}\right)\approx1.148$, option (B).
The line just touches the curve: it’s the tangent. Move the point anywhere and $3\cos(3x)$ still gives the curve’s gradient.
This is a point of inflection, so the tangent crosses the curve here instead of just touching it. It still has the curve’s gradient, and $3\cos(3x)$ gives that gradient wherever you put the point.
Here $3\sin(3x)$ happens to give the tangent too, because $\sin(3x)=\cos(3x)$ at this point. Move the point and it stops matching.
The curve is flat here, so its gradient is 0, and $\cos(3x)$ is 0 too. Move the point and it stops matching.
The blue line cuts across the curve instead of just touching it. $3\sin(3x)$ gives 2.772 here, option (A), but the curve’s gradient is only about 1.148.
The blue line cuts across the curve instead of just touching it, so $3\sin(3x)$ isn’t the gradient here.
The blue line is too shallow. $\cos(3x)$ gives 0.383 here, option (D), which is only a third of the curve’s gradient. The chain rule’s factor of 3 is missing.
The blue line cuts across the curve. $\cos(3x)$ is always a third of the true gradient, because it’s missing the chain rule’s factor of 3.
The angle is $3x$, not just $x$. So $\sin(3x)$ is a function inside a function.
$$f(x)=\sin(u),\qquad u=3x$$
The chain rule multiplies the derivative of the outside function by the derivative of the inside one.
$\displaystyle \frac{d}{dx}\sin\big(g(x)\big)$$\displaystyle =g'(x)\cos\big(g(x)\big)$
Here $g(x)=3x$, so $g'(x)=3$.
$$f'(x)=3\cos(3x)$$
The angle $\tfrac{\pi}{8}$ is in radians, so put your calculator in radian mode first.
$$f'\left(\tfrac{\pi}{8}\right)=3\cos\left(\tfrac{3\pi}{8}\right)\approx1.148$$
In the figure, the line with gradient $3\cos(3x)$ is the tangent wherever you put the point.
$$\begin{aligned}f'(x)&=3\cos(3x)\\f'\left(\tfrac{\pi}{8}\right)&=3\cos\left(\tfrac{3\pi}{8}\right)\\&\approx1.148\end{aligned}$$
Differentiate with the chain rule, then evaluate with your calculator in radian mode.
How Queensland students answered in 2023
From the QCAA Mathematical Methods subject report for 2023. The percentages can add to less than 100 because some students didn’t answer.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording is from the 2023 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2023, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
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