Question 15
Determine the derivative of $f(x)=\ln x^2+\ln(x-5)^3$. Express the derivative as a single fraction in its simplest and factorised form.
[4 marks]Pull the powers out, then differentiate
A log law turns each power into a coefficient. Then each term differentiates in one line, and the rest is fraction work.
With the powers out the front, each log term differentiates to a simple fraction. The last two marks are all about the form of the answer.
Right gradient, right form?
An answer has to pass two tests. Its value must be the gradient of the curve, and its form must be what the question asks for. The orange line is the true tangent. Choose an expression to draw its line dashed, then move along the curve.
Choose an expression. A correct derivative draws its dashed line exactly on top of the orange tangent, wherever you move along the curve.
Right derivative: the dashed line sits on the tangent for every $x$. But it is two fractions, and the question asks for a single fraction. This matches the first two marks. The last two need one fraction, then a factorised one.
One fraction, and the dashed line still sits on the tangent for every $x$. But $5x-10$ has a common factor of 5, so the numerator isn’t factorised yet.
This is it. A single fraction, with the numerator and the denominator both factorised and nothing left to cancel. The dashed line sits on the tangent for every $x$.
This misses the chain rule. The derivative of $\ln u$ is $\frac{u'}{u}$, so $\ln x^2$ gives $\frac{2x}{x^2}=\frac{2}{x}$, not $\frac{1}{x^2}$.
Work through the solution one mark at a time
Try each step yourself before you reveal it. This is a by-hand question even on the technology-active paper, because a calculator won’t give you the factorised form. Use it to check your answer instead.
Before differentiating, use the log law $\ln\left(a^n\right)=n\ln a$ on both terms. That turns each one into a number times a plain log.
$$\begin{aligned}f(x)&=\ln x^2+\ln(x-5)^3\\f(x)&=2\ln x+3\ln(x-5)\end{aligned}$$
The derivative of $\ln x$ is $\dfrac{1}{x}$, so the first term gives:
$$\frac{d}{dx}\left(2\ln x\right)=2\times\frac{1}{x}=\frac{2}{x}$$
Equivalent formats are accepted. If you skip the log law and use the chain rule on $\ln x^2$ directly, you get $\dfrac{2x}{x^2}$, which is the same thing.
Marker: correctly determines first term derivative
For the second term, the inside is $x-5$, whose derivative is 1. So the chain rule changes nothing here.
$$\begin{aligned}&\frac{d}{dx}\left(3\ln(x-5)\right)\\&=3\times\frac{1}{x-5}\\&=\frac{3}{x-5}\end{aligned}$$
So far, $f'(x)=\dfrac{2}{x}+\dfrac{3}{x-5}$. Equivalent formats are accepted for this mark too, such as $\dfrac{3(x-5)^2}{(x-5)^3}$ from the chain rule without the log law.
Marker: correctly determines second term derivative
“A single fraction” means one numerator over one denominator. The common denominator is $x(x-5)$. Multiply each numerator by whatever its own denominator is missing.
$$f'(x)=\frac{2(x-5)+3x}{x(x-5)}$$
Follow-through applies, so if a derivative in the first two steps had a slip, combining your own fractions correctly can still earn this mark.
Marker: combines the fractions to one fraction
Expand the numerator and collect like terms. Then take out the common factor of 5. The denominator is already factorised, so leave it as it is rather than expanding it.
$$\begin{aligned}f'(x)&=\frac{2x-10+3x}{x(x-5)}\\&=\frac{5x-10}{x(x-5)}\end{aligned}$$
Check it on your calculator. Use the derivative-at-a-point function on the original $f(x)$ at a value such as $x=6$. It gives 3.3333…, and your fraction gives $\dfrac{5(4)}{6(1)}=\dfrac{10}{3}$, so the algebra checks out. Pick a value bigger than 5, because the function only exists for $x>5$.
Marker: determines the simplest and factorised form
Each phrase of the question gave you something
Differentiating logs and adding fractions are both routine. The log law makes the calculus a one-liner for each term, and everything after that is the same fraction work you did in junior maths, just with $x$ in the denominators.
If I had to guess where you might lose marks, it would be stopping early. Half the marks are for the form, so $\tfrac{2}{x}+\tfrac{3}{x-5}$ is only halfway, and $\tfrac{5x-10}{x(x-5)}$ is one step short. The other risk is the chain rule. Without it, $\ln x^2$ becomes $\tfrac{1}{x^2}$, which is wrong. Using the log law first avoids that trap completely.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording and marking-guide steps are from the 2023 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2023, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.
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