Unit 3 · Topic 2.2 · Differentiation of trigonometric functions and differentiation rules
You have already met the chain, product and quotient rules in the last three guides. In the external they get applied to exponentials, logs and trig all the time, so I have given them their own guide here. Tick off each skill once you can do it without help.
Try it yourself
Before you differentiate anything, decide which rule the function needs. Most mistakes I see in this topic come from picking the wrong one. The algebra is rarely the problem. Choose a rule for each function below.
Part 1
Use the chain rule when one function sits inside another, like $\big(f(x)\big)^{n}$. The formal version substitutes $u$ for the inside function. The fast version does the same thing in one line.
The fast way is very efficient, but keep the formal version in your head as well for the really tricky derivatives. If you also take Specialist, you need to understand both.
Watch the fast version
Practise
For the written-answer questions, the solutions show where each mark goes, based on the QCAA marking guides. Many marks allow follow-through: if an earlier step is wrong but you use your wrong answer correctly, you can still get the later marks. That is another reason to show every step.
If $f(x)=e^{3x}(x+1)^{2}$ and $f'(x)=ae^{3x}(x+1)$, determine the expression for $a$.
It is a product, and each part needs the chain rule. Let $u=e^{3x}$ and $v=(x+1)^{2}$, so $u'=3e^{3x}$ and $v'=2(x+1)$:
$$f'(x)=3e^{3x}(x+1)^{2}+2e^{3x}(x+1)$$
Take out the common factor $e^{3x}(x+1)$. What is left is $3(x+1)+2=3x+5$.
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.
You could use the chain rule on each log, but the log laws make it much quicker. Bring the powers down first:
$$f(x)=2\ln x+3\ln(x-5)$$
Now differentiate each term.
$$f'(x)=\frac{2}{x}+\frac{3}{x-5}$$
Now make it a single fraction:
$$f'(x)=\frac{2(x-5)+3x}{x(x-5)}=\frac{5x-10}{x(x-5)}$$
Then factorise the top.
Determine the derivative with respect to $x$ of $y=\left(e^{x}+1\right)^{3}$.
This is a function raised to a power, so use the chain rule, with $f(x)=e^{x}+1$ and $n=3$. Bring the power down, multiply by the derivative of the inside, then drop the power by one.
$$\frac{dy}{dx}=3\times e^{x}\times\left(e^{x}+1\right)^{2}$$
Determine the derivative with respect to $x$ of $y=\dfrac{1}{\sin(x)}$.
It looks like a quotient, but the top is just 1. Rewrite it as a power and the chain rule is faster: $y=(\sin x)^{-1}$.
$$\frac{dy}{dx}=-1\times\cos x\times(\sin x)^{-2}$$
Part 2
Use the product rule when two functions are multiplied together. Your formula sheet writes it one way, and we have mostly used the other in class. They mean exactly the same thing, but make sure you recognise both before the exam.
For the product rule, it does not matter which function you call $u$ and which you call $v$. The two terms are added, so the order makes no difference.
Watch it first
Divide the extra area by the small change in $x$ and let it shrink to zero. The two strips become $vu'$ and $uv'$. That is why the product rule has two terms, one for each function changing.
Practise
Differentiate $y=\ln(x)\cos(x)$ with respect to $x$.
Let $u=\ln x$ and $v=\cos x$, so $u'=\frac{1}{x}$ and $v'=-\sin x$. That minus sign is the whole difference between (C) and (D).
$$\frac{dy}{dx}=\ln x\,(-\sin x)+\cos x\cdot\frac{1}{x}$$
Determine the derivative of $y=2x\cos(3x)$.
Let $u=2x$ and $v=\cos(3x)$, so $u'=2$ and, with the chain rule, $v'=-3\sin(3x)$. (C) differentiates each part and multiplies them, which is not the product rule.
$$\frac{dy}{dx}=2x\left(-3\sin(3x)\right)+\cos(3x)\cdot 2$$
Determine the second derivative of $h(x)=x\sin(x)$. Give your answer in simplest form.
You need the product rule twice. First, with $u=x$ and $v=\sin x$:
$$h'(x)=x\cos x+\sin x$$
Now differentiate again. The first term is another product, and the second is just $\cos x$. You differentiate a sum one term at a time.
$$h''(x)=\big(x(-\sin x)+\cos x\big)+\cos x$$
Collect the two $\cos x$ terms for simplest form.
The equation of the tangent to the curve $f(t)=te^{t}$ at $t=1$ is
A tangent needs a point and a gradient. The point is $f(1)=e$. For the gradient, use the product rule with $u=t$ and $v=e^{t}$:
$$f'(t)=te^{t}+e^{t},\qquad f'(1)=2e$$
Then $y-e=2e(t-1)$, which tidies up to $y=2et-e$.
Determine the derivative with respect to $x$ of $y=x^{2}\times e^{-x}$. Express your answer in factorised form.
Let $u=x^{2}$ and $v=e^{-x}$, so $u'=2x$ and $v'=-e^{-x}$:
$$\frac{dy}{dx}=x^{2}\left(-e^{-x}\right)+e^{-x}\cdot 2x$$
The question asks for factorised form, so take out $xe^{-x}$. Stopping before this step loses you a mark.
Part 3
Use the quotient rule when one function is divided by another. Again, the formula sheet and our class version look different but mean the same thing.
Unlike the product rule, here it matters which function is which, because of the subtraction. $v$ is always the denominator of the original expression. Get that right and the rest follows.
Watch the top get built
Practise
Given that $g(x)=\dfrac{\ln(x)}{x}$, determine the simplest value of $g'(e)$.
It is one function divided by another, so use the quotient rule. The denominator is $x$, so $v=x$ and $u=\ln x$. Then $u'=\frac{1}{x}$ and $v'=1$:
$$g'(x)=\frac{x\cdot\frac{1}{x}-\ln x\cdot 1}{x^{2}}=\frac{1-\ln x}{x^{2}}$$
Now substitute $x=e$. Since $\ln e=1$, the numerator is $0$.
$$g'(e)=\frac{1-\ln e}{e^{2}}=\frac{0}{e^{2}}=0$$
Determine the derivative with respect to $x$ of $y=\dfrac{\sin(x)}{x^{2}}$. Give your answer in simplest form.
It is one function divided by another, so use the quotient rule. The denominator is $x^{2}$, so $v=x^{2}$ and $u=\sin x$:
$$\frac{dy}{dx}=\frac{x^{2}\cos x-2x\sin x}{x^{4}}$$
"Simplest form" is worth a mark here. Every term on top has a factor of $x$, so cancel one $x$ from the top and bottom.
Game plan
Ask yourself these, in order, before you start differentiating. Try them on the functions in the sorter at the top.
Practice
Eight questions of my own, covering every skill. Each one is labelled with the skill it practises, so if one goes wrong you know which part of the guide to go back to. For the first six, the rule is in the label. For the last two, deciding which rule goes on the outside is part of the question.
Determine $f'(x)$ for $f(x)=\sqrt{e^{2x}+1}$, giving your answer in simplest form.
$$f'(x)=\frac{1}{2}\times 2e^{2x}\times\left(e^{2x}+1\right)^{-\frac{1}{2}}$$
The derivative of $\ln(\cos x)$ with respect to $x$ is
The inside is $\cos x$, and its derivative is $-\sin x$. The rule for $\ln f(x)$ gives $\dfrac{f'(x)}{f(x)}=\dfrac{-\sin x}{\cos x}=-\tan x$. Option (A) has forgotten the chain rule, and (C) has lost the minus sign from differentiating $\cos x$.
Determine $\dfrac{dy}{dx}$ for $y=x^{3}\ln x$, giving your answer in factorised form.
$$\frac{dy}{dx}=x^{3}\times\frac{1}{x}+\ln x\times 3x^{2}=x^{2}+3x^{2}\ln x$$
Determine the exact gradient of the curve $y=e^{-2x}\cos x$ at $x=\dfrac{\pi}{2}$.
$$\frac{dy}{dx}=e^{-2x}(-\sin x)+\cos x\left(-2e^{-2x}\right)=-e^{-2x}\left(\sin x+2\cos x\right)$$
$$\frac{dy}{dx}=-e^{-\pi}\left(1+0\right)=-e^{-\pi}$$
Factorising first makes the substitution much tidier. It also means only one exponential to evaluate.
Let $y=\dfrac{\ln x}{x}$ for $x>0$. Show that $\dfrac{dy}{dx}=\dfrac{1-\ln x}{x^{2}}$, then determine the exact coordinates of the point where the tangent to the curve is horizontal.
$$\frac{dy}{dx}=\frac{x\times\frac{1}{x}-\ln x\times 1}{x^{2}}=\frac{1-\ln x}{x^{2}}$$
That matches the expression in the question. In a "show that", the working is the answer, so write every line.
$$1-\ln x=0\quad\Rightarrow\quad \ln x=1\quad\Rightarrow\quad x=e$$
$$y=\frac{\ln e}{e}=\frac{1}{e}$$
If $y=\dfrac{e^{x}}{e^{x}+1}$, then $\dfrac{dy}{dx}$ is
Here $u=e^{x}$ and $v=e^{x}+1$, and both derivatives are $e^{x}$. The top is $vu'-uv'=\left(e^{x}+1\right)e^{x}-e^{x}\times e^{x}=e^{x}$, so the answer is $\dfrac{e^{x}}{\left(e^{x}+1\right)^{2}}$. Option (B) has the subtraction the wrong way round, (C) divides the derivatives instead of using the rule, and (D) adds instead of subtracting.
Determine $\dfrac{dy}{dx}$ for $y=xe^{x^{2}}$, giving your answer in factorised form.
$$\frac{dy}{dx}=x\times 2xe^{x^{2}}+e^{x^{2}}\times 1=2x^{2}e^{x^{2}}+e^{x^{2}}$$
The concentration of a medicine in a patient's blood is modelled by $C(t)=20te^{-0.4t}$, where $C$ is in mg/L and $t$ is the time in hours after the dose.
(a) Determine the rate at which the concentration is changing 3 hours after the dose. [2 marks]
(b) Determine when the concentration is greatest, and the greatest concentration. [3 marks]
$$C'(t)=20t\left(-0.4e^{-0.4t}\right)+e^{-0.4t}\times 20=e^{-0.4t}\left(20-8t\right)$$
$$C'(3)=e^{-1.2}\left(20-24\right)\approx -1.20$$
The negative sign matters. It tells you the concentration is falling at this time, so say so.
$$C(2.5)=20\times 2.5\times e^{-1}\approx 18.39$$
Keep practising
In 2021 Paper 2, Question 18 you differentiate a logistic model, which is a quotient with a chain rule inside. The walkthrough lets you drag a tangent along the curve to find where the growth peaks.
Past questions and marking-guide steps are from the 2020–2024 QCAA Mathematical Methods external assessments, © State of Queensland (QCAA) 2020–2024, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.
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