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Differentiation rules and standard derivatives: VCE Math Methods Unit 3

Syllabus dot point

“The product, quotient and chain rules of differentiation, and the derivatives of standard functions xnx^n for n∈Qn \in Q, exe^x, ln⁡(x)\ln(x), sin⁡(x)\sin(x), cos⁡(x)\cos(x) and tan⁡(x)\tan(x)”

9 min read

Quick answer

The four differentiation rules (sum, product, quotient, chain) combined with the standard derivatives of xnx^n, exe^x, ln⁡(x)\ln(x), sin⁡(x)\sin(x), cos⁡(x)\cos(x) and tan⁡(x)\tan(x) let you differentiate any function in the VCE Math Methods Paper 1 library, with the chain rule the single most-tested rule and the product and quotient rules requiring careful attention to signs.

Jump to a section
  1. What this dot point is asking
  2. Standard derivatives
  3. The sum rule
  4. The product rule
  5. The quotient rule
  6. The chain rule
  7. Combining rules
  8. Examples in context
  9. Try this
  10. Exam-style questions

What this dot point is asking

VCAA wants fluent by-hand differentiation of any function built from the standard library (polynomials, exe^x, ln⁡x\ln x, sin⁡x\sin x, cos⁡x\cos x, tan⁡x\tan x) using the four standard rules. Paper 1 almost always opens with a differentiation question and rewards clean, factored answers.

Standard derivatives

Memorise these. They appear in nearly every paper.

function derivative
xnx^n for n∈Qn \in \mathbb{Q} nxn−1n x^{n-1}
exe^x exe^x
ekxe^{k x} kekxk e^{k x}
ln⁡(x)\ln(x) 1x\frac{1}{x}
sin⁡(x)\sin(x) cos⁡(x)\cos(x)
cos⁡(x)\cos(x) −sin⁡(x)-\sin(x)
tan⁡(x)\tan(x) sec⁡2(x)=1cos⁡2(x)\sec^2(x) = \frac{1}{\cos^2(x)}

The power rule ddx(xn)=nxn−1\frac{d}{dx}(x^n) = n x^{n-1} extends to all rational nn: e.g. ddx(x)=12x−1/2\frac{d}{dx}(\sqrt{x}) = \frac{1}{2} x^{-1/2}, and ddx(x−2)=−2x−3\frac{d}{dx}(x^{-2}) = -2 x^{-3}.

The derivative of ln⁡(x)\ln(x) is 1x\frac{1}{x} for x>0x > 0. For the extended form, ddx(ln⁡∣x∣)=1x\frac{d}{dx}(\ln|x|) = \frac{1}{x} for x≠0x \neq 0.

The sum rule

ddx[f(x)+g(x)]=f′(x)+g′(x)\frac{d}{dx}[f(x) + g(x)] = f'(x) + g'(x)

You differentiate term by term. Constants come out: ddx[c⋅f(x)]=c⋅f′(x)\frac{d}{dx}[c \cdot f(x)] = c \cdot f'(x).

The product rule

If y=u(x)v(x)y = u(x) v(x), then

dydx=u′(x)v(x)+u(x)v′(x)\frac{dy}{dx} = u'(x) v(x) + u(x) v'(x)

In words: derivative of the first times the second, plus the first times derivative of the second.

Example. Differentiate y=x2exy = x^2 e^x.

u=x2u = x^2, v=exv = e^x. u′=2xu' = 2 x, v′=exv' = e^x.

dydx=2x⋅ex+x2⋅ex=ex(2x+x2)=xex(2+x)\frac{dy}{dx} = 2 x \cdot e^x + x^2 \cdot e^x = e^x (2 x + x^2) = x e^x (2 + x).

The quotient rule

If y=u(x)v(x)y = \frac{u(x)}{v(x)}, then

dydx=u′(x)v(x)−u(x)v′(x)[v(x)]2\frac{dy}{dx} = \frac{u'(x) v(x) - u(x) v'(x)}{[v(x)]^2}

Note the sign: u′vu' v minus uv′u v' in that order.

Example. Differentiate y=xln⁡(x)y = \frac{x}{\ln(x)} for x>0x > 0, x≠1x \neq 1.

u=xu = x, v=ln⁡(x)v = \ln(x). u′=1u' = 1, v′=1xv' = \frac{1}{x}.

dydx=1⋅ln⁡(x)−x⋅1x[ln⁡(x)]2=ln⁡(x)−1[ln⁡(x)]2\frac{dy}{dx} = \frac{1 \cdot \ln(x) - x \cdot \frac{1}{x}}{[\ln(x)]^2} = \frac{\ln(x) - 1}{[\ln(x)]^2}.

The chain rule

If y=f(g(x))y = f(g(x)), let u=g(x)u = g(x) so y=f(u)y = f(u). Then

dydx=dydu⋅dudx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

In words: differentiate the outside leaving the inside alone, then multiply by the derivative of the inside.

Example. Differentiate y=sin⁡(3x2)y = \sin(3 x^2).

Inside: u=3x2u = 3 x^2, dudx=6x\frac{du}{dx} = 6 x. Outside: dydu=cos⁡(u)\frac{dy}{du} = \cos(u).

dydx=cos⁡(3x2)⋅6x=6xcos⁡(3x2)\frac{dy}{dx} = \cos(3 x^2) \cdot 6 x = 6 x \cos(3 x^2).

Chain rule shortcuts for common composites

These are worth memorising as patterns:

  • ddx(ef(x))=f′(x)ef(x)\frac{d}{dx}(e^{f(x)}) = f'(x) e^{f(x)}
  • ddx(ln⁡(f(x)))=f′(x)f(x)\frac{d}{dx}(\ln(f(x))) = \frac{f'(x)}{f(x)} for f(x)>0f(x) > 0
  • ddx(sin⁡(f(x)))=f′(x)cos⁡(f(x))\frac{d}{dx}(\sin(f(x))) = f'(x) \cos(f(x))
  • ddx(cos⁡(f(x)))=−f′(x)sin⁡(f(x))\frac{d}{dx}(\cos(f(x))) = -f'(x) \sin(f(x))
  • ddx([f(x)]n)=n[f(x)]n−1f′(x)\frac{d}{dx}([f(x)]^n) = n [f(x)]^{n-1} f'(x)

Combining rules

Many Paper 1 questions combine two or three rules.

Worked example. Differentiate f(x)=x2e3xf(x) = x^2 e^{3 x}.

Product rule with u=x2u = x^2, v=e3xv = e^{3 x}. u′=2xu' = 2 x, v′=3e3xv' = 3 e^{3 x} (chain rule on the inside).

f′(x)=2x⋅e3x+x2⋅3e3x=e3x(2x+3x2)=xe3x(2+3x)f'(x) = 2 x \cdot e^{3 x} + x^2 \cdot 3 e^{3 x} = e^{3 x} (2 x + 3 x^2) = x e^{3 x} (2 + 3 x).

Worked example. Differentiate f(x)=ln⁡(cos⁡(x))f(x) = \ln(\cos(x)).

Chain rule with outside ln⁡\ln and inside cos⁡(x)\cos(x).

f′(x)=1cos⁡(x)⋅(−sin⁡(x))=−sin⁡(x)cos⁡(x)=−tan⁡(x)f'(x) = \frac{1}{\cos(x)} \cdot (-\sin(x)) = -\frac{\sin(x)}{\cos(x)} = -\tan(x).

Common Paper 1 traps
Forgetting the chain rule on composed functions
Writing ddx(sin⁡(2x))=cos⁡(2x)\frac{d}{dx}(\sin(2x)) = \cos(2x) loses the factor of 22. Correct: 2cos⁡(2x)2 \cos(2x).
Sign error in the quotient rule
The numerator is u′vu' v minus uv′u v', in that order. Reversing it flips the sign.
Power rule on axa^x for a≠ea \neq e
ddx(2x)\frac{d}{dx}(2^x) is not x⋅2x−1x \cdot 2^{x - 1}. Use 2x=exln⁡22^x = e^{x \ln 2} and the chain rule: ddx(2x)=(ln⁡2)⋅2x\frac{d}{dx}(2^x) = (\ln 2) \cdot 2^x.

Treating e2xe^{2x} as 2e2x2 e^{2x} without the chain rule. The chain rule gives ddx(e2x)=2e2x\frac{d}{dx}(e^{2x}) = 2 e^{2x}, but the working should show the chain rule explicitly.

Stopping before simplification. Markers usually reward a clean factored form. After the quotient rule, look for common factors in numerator and denominator.

Examples in context

Example 1. Rate of change of a damped oscillation. A spring's displacement is s(t)=e−tsin⁡(t)s(t) = e^{-t}\sin(t). Using the product rule with u=e−tu = e^{-t} (u′=−e−tu' = -e^{-t}) and v=sin⁡tv = \sin t (v′=cos⁡tv' = \cos t): s′(t)=−e−tsin⁡t+e−tcos⁡t=e−t(cos⁡t−sin⁡t)s'(t) = -e^{-t}\sin t + e^{-t}\cos t = e^{-t}(\cos t - \sin t). The velocity is zero when cos⁡t=sin⁡t\cos t = \sin t, i.e. tan⁡t=1\tan t = 1, first at t=π4t = \frac{\pi}{4}.

Example 2. Marginal cost from a power model. A firm's cost is C(x)=50x=50x1/2C(x) = 50\sqrt{x} = 50x^{1/2} thousand dollars for xx hundred units. The marginal cost is C′(x)=50×12x−1/2=25xC'(x) = 50 \times \frac{1}{2}x^{-1/2} = \frac{25}{\sqrt{x}}. At x=25x = 25, C′(25)=255=5C'(25) = \frac{25}{5} = 5 thousand dollars per extra hundred units, showing the rational power rule in use.

Try this

Q1. Differentiate y=(2x+1)5y = (2x + 1)^5. [2 marks]

  • Cue. Chain rule: y′=5(2x+1)4×2=10(2x+1)4y' = 5(2x + 1)^4 \times 2 = 10(2x + 1)^4.

Q2. Differentiate f(x)=x2sin⁡xf(x) = x^2 \sin x. [3 marks]

  • Cue. Product rule: f′(x)=2xsin⁡x+x2cos⁡xf'(x) = 2x\sin x + x^2\cos x.

Q3. Differentiate y=exxy = \dfrac{e^x}{x}. [3 marks]

  • Cue. Quotient rule: y′=ex⋅x−ex⋅1x2=ex(x−1)x2y' = \dfrac{e^x \cdot x - e^x \cdot 1}{x^2} = \dfrac{e^x(x - 1)}{x^2}.

Exam-style questions

Questions in the style of VCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

2023 VCAA-style (Exam 1)3 marks
Differentiate f(x)=x3ln⁡(x)f(x) = x^3 \ln(x) with respect to xx.
Show worked answer →

Use the product rule with u=x3u = x^3 and v=ln⁡(x)v = \ln(x).

u′=3x2u' = 3 x^2 and v′=1xv' = \frac{1}{x}.

f′(x)=u′v+uv′=3x2ln⁡(x)+x3⋅1x=3x2ln⁡(x)+x2f'(x) = u' v + u v' = 3 x^2 \ln(x) + x^3 \cdot \frac{1}{x} = 3 x^2 \ln(x) + x^2.

Factor: f′(x)=x2(3ln⁡(x)+1)f'(x) = x^2 (3 \ln(x) + 1).

Markers reward explicit labelling of uu, vv, u′u' and v′v', correct application of the product rule, and a tidy final answer.

2024 VCAA-style (Exam 1)3 marks
Differentiate y=sin⁡(2x)exy = \frac{\sin(2x)}{e^x} with respect to xx.
Show worked answer →

Quotient rule with u=sin⁡(2x)u = \sin(2x), v=exv = e^x.

u′=2cos⁡(2x)u' = 2 \cos(2x) (chain rule on the inside), v′=exv' = e^x.

dydx=u′v−uv′v2=2cos⁡(2x)⋅ex−sin⁡(2x)⋅ex(ex)2=ex(2cos⁡(2x)−sin⁡(2x))e2x=2cos⁡(2x)−sin⁡(2x)ex\frac{dy}{dx} = \frac{u' v - u v'}{v^2} = \frac{2 \cos(2x) \cdot e^x - \sin(2x) \cdot e^x}{(e^x)^2} = \frac{e^x (2 \cos(2x) - \sin(2x))}{e^{2x}} = \frac{2 \cos(2x) - \sin(2x)}{e^x}.

Markers reward the chain rule inside the quotient rule, correct sign, and simplification to remove the common exe^x factor.

Practise this

Sources & how we know this

ExamExplained