WJEC Unit 1 Specimen — Question 3 6 marks

Exam BoardWJEC
ModuleUnit 1 (Unit 1)
SessionSpecimen
Marks6
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TopicDifferentiation from First Principles
TypeFirst principles: x³ terms
DifficultyModerate -0.5 This is a standard first principles differentiation question requiring students to apply the definition of the derivative using the limit of (f(x+h)-f(x))/h. While it involves algebraic manipulation of (x+h)³ and cancelling h, it's a routine textbook exercise testing a fundamental technique with no problem-solving required. The 6 marks reflect the working steps rather than conceptual difficulty, making it slightly easier than average.
Spec1.07g Differentiation from first principles: for small positive integer powers of x
Given that \(y = x^3\), find \(\frac{dy}{dx}\) from first principles. [6]
AnswerMarks Guidance
\(y + k = (x + h)^3\)M1 AO2
\(y + k = x^3 + 3x^2h + 3xh^2 + h^3\)A1 AO2
Subtracting \(y\) from above to find \(k\)M1 AO2
\(k = 3x^2h + 3xh^2 + h^3\)A1 AO2
Dividing by \(h\) and letting \(h \to 0\)M1 AO2
\(\frac{dy}{dx} = \lim_{h \to 0} \frac{k}{h} = 3x^2\)A1 AO2 (c.a.o.)
Total: [6]
$y + k = (x + h)^3$ | M1 | AO2
$y + k = x^3 + 3x^2h + 3xh^2 + h^3$ | A1 | AO2
Subtracting $y$ from above to find $k$ | M1 | AO2
$k = 3x^2h + 3xh^2 + h^3$ | A1 | AO2
Dividing by $h$ and letting $h \to 0$ | M1 | AO2
$\frac{dy}{dx} = \lim_{h \to 0} \frac{k}{h} = 3x^2$ | A1 | AO2 (c.a.o.)

**Total: [6]**

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Given that $y = x^3$, find $\frac{dy}{dx}$ from first principles. [6]

\hfill \mbox{\textit{WJEC Unit 1  Q3 [6]}}
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