2018 THREE(d)

Merit
Question
Find the equation of the tangent to the curve y=x2ln⁡xy = x^2\ln x at the point where x=ex = e.

*You must use calculus and show any derivatives that you need to find when solving this problem.*
Official Answer
dydx=2x⋅ln⁡x+x2⋅1x\dfrac{dy}{dx}=2x\cdot \ln x + x^2\cdot \dfrac{1}{x}

dydx=2x⋅ln⁡x+x\dfrac{dy}{dx}=2x\cdot \ln x + x

x=e⇒dydx=2e⋅ln⁡e+ex=e\Rightarrow \dfrac{dy}{dx}=2e\cdot \ln e + e

=3e=3e

Equation of tangent:

y−y1=m(x−x1)y-y_1=m(x-x_1)

y−e2=3e(x−e)y-e^2=3e(x-e)

y−e2=3ex−3e2y-e^2=3ex-3e^2

y=3ex−2e2y=3ex-2e^2

(y=8.155x−14.778)(y=8.155x-14.778)
Grading Criteria

Achievement (u)

  • Correct expression for dydx\dfrac{dy}{dx}

Merit (r)

  • Correct solution with correct derivative.
  • Accept any equivalent form.

Excellence T1

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Excellence T2

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Video Explanation
NCEA Level 3 Calculus Differentiation 2018 NZQA Exam - Worked Answers by infinityplusone(starts at 48:00)
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Exam Paper (2018)Assessment Schedule (2018)
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