2016 TWO(b)

Achievement
Question
Find the gradient of the tangent to the function y=2x−1y = \sqrt{2x - 1} at the point (5,3)(5, 3).

*You must use calculus and show any derivatives that you need to find when solving this problem.*
Official Answer
y=(2x−1)12y=(2x-1)^{\frac{1}{2}}
dydx=12(2x−1)−12⋅2\frac{dy}{dx}=\frac{1}{2}(2x-1)^{-\frac{1}{2}}\cdot 2
=12x−1=\frac{1}{\sqrt{2x-1}}
At x=5x=5, dydx=13\frac{dy}{dx}=\frac{1}{3}
Grading Criteria

Achievement (u)

  • Correct solution with correct derivative.

Merit (r)

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Excellence T1

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Excellence T2

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Exam Paper (2016)Assessment Schedule (2016)
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