NCEA Calculus 2019 TWO(e - Solution & Explanation

Question

If

y = e^u

and

u = \sin 2x

show that

\dfrac{d^2 y}{dx^2} = \dfrac{d^2 y}{du^2}\left(\dfrac{du}{dx}\right)^2 + \dfrac{dy}{du}\cdot\dfrac{d^2 u}{dx^2}

*You must use calculus and show any derivatives that you need to find when solving this
problem.*

Official Answer

LHS

y=e^{\sin 2x}

\dfrac{dy}{dx}=e^{\sin 2x}\times 2\cos 2x

\dfrac{d^2y}{dx^2}=e^{\sin 2x}\times(-4\sin 2x)+e^{\sin 2x}\times(2\cos 2x)^2

u=\sin 2x

\dfrac{du}{dx}=2\cos 2x

\dfrac{d^2u}{dx^2}=-4\sin 2x

y=e^u

\dfrac{dy}{du}=e^u

\dfrac{d^2y}{du^2}=e^u

RHS

\dfrac{d^2y}{du^2}\times\left(\dfrac{du}{dx}\right)^2+\dfrac{dy}{du}\times\dfrac{d^2u}{dx^2}

=e^u\times(2\cos 2x)^2+e^u\times(-4\sin 2x)

=e^{\sin 2x}\times(2\cos 2x)^2+e^{\sin 2x}\times(-4\sin 2x)

Therefore LHS = RHS as required.

\dfrac{d^2y}{dx^2}=4e^{\sin 2x}(\cos^2 2x-\sin 2x)

Grading Criteria

Correct expressions for $\dfrac{d^2y}{dx^2}$ in any equivalent form.
Or correct RHS.

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Video Explanation

NCEA Level 3 Calculus Differentiation 2019 NZQA Exam - Worked Answers by infinityplusone(starts at 36:58)

2019 TWO(e)