Moderate -0.8 This is a straightforward application of the first principles definition to differentiate a simple quadratic. While it requires careful algebraic manipulation of the limit definition, it's a standard textbook exercise with a single power term and no complications. The 'show that' format makes it slightly easier than finding the derivative independently.
Uses the given function in the formula. Allow a slip e.g. missing brackets
\(= \frac{2x^2+4xh+2h^2+3-(2x^2+3)}{h}\)
M1
Attempt to simplify the numerator
\(= 4x+2h\)
A1
Correct expression without \(h\) in the denominator from fully correct working
\(f'(x) = \lim_{h \to 0}(4x+2h) = 4x\)
A1
AG Correct use of limit as \(h \to 0\) with their expression
[4]
## Question 6:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{f(x+h)-f(x)}{h} = \frac{2(x+h)^2+3-(2x^2+3)}{h}$ | M1 | Uses the given function in the formula. Allow a slip e.g. missing brackets |
| $= \frac{2x^2+4xh+2h^2+3-(2x^2+3)}{h}$ | M1 | Attempt to simplify the numerator |
| $= 4x+2h$ | A1 | Correct expression without $h$ in the denominator from fully correct working |
| $f'(x) = \lim_{h \to 0}(4x+2h) = 4x$ | A1 | AG Correct use of limit as $h \to 0$ with their expression |
| [4] | | |
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6 Given that $\mathrm { f } ( x ) = 2 x ^ { 2 } + 3$, show from first principles that $\mathrm { f } ^ { \prime } ( x ) = 4 x$.
\hfill \mbox{\textit{OCR MEI Paper 1 2024 Q6 [4]}}