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Before the final

Evaluating the derivative at the wrong point

The habit: The derivative is taken at the base point, the point of tangency on the curve; find that point before substituting.

14 practice questions are built to catch it, in 8 topics.

Practise against it: 6 questions

What it looks like

A question from linear approximation, with the answer this slip gives beside the right one.

Introductory1 mark · 2 min

Use a linear approximation of f(x)=xf(x) = \sqrt{x} at x=25x = 25 to estimate 26\sqrt{26}.

With the slip
≈5.098\displaystyle \approx 5.098
Worked through
26≈5.1\displaystyle \sqrt{26} \approx 5.1

1/(2√26) is the slope at 26, which needs the very root being estimated. The line is built at 25, from f(25) and f′(25).

See the worked solution

Where it comes up

Practise against it

Each of these has an answer this slip gives. Work them on paper: if you fall into it, the verdict names it and the question comes back in review.

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