Question:

<b>Implicit Differentiation after Substitution</b><br> Let $y=\sin\alpha$, $x=\sin\beta$. Then the equation becomes:<br> $\sin\alpha\cos\beta = k-\sin\beta\cos\alpha\Rightarrow\sin(\alpha+\beta)=k$.<br> So $\alpha+\beta=\sin^{-1}k=$ constant, meaning $\sin^{-1}y+\sin^{-1}x=C$.<br> Differentiate: $\dfrac{1}{\sqrt{1-y^2}}\dfrac{dy}{dx}+\dfrac{1}{\sqrt{1-x^2}}=0\Rightarrow\dfrac{dy}{dx}=-\dfrac{\sqrt{1-y^2}}{\sqrt{1-x^2}}$.<br> At $x=\tfrac{1}{2}$, $y=-\tfrac{1}{4}$: $1-y^2=1-\tfrac{1}{16}=\tfrac{15}{16}$, $1-x^2=\tfrac{3}{4}$.<br> $\dfrac{dy}{dx}=-\dfrac{\sqrt{15/16}}{\sqrt{3/4}}=-\dfrac{\sqrt{15}/4}{\sqrt{3}/2}=-\dfrac{\sqrt{15}}{4}\cdot\dfrac{2}{\sqrt{3}}=-\dfrac{\sqrt{5}}{2}$.<br> Option (2) is $\sqrt{5}/2$... sign: $\dfrac{dy}{dx}=-\sqrt{5}/2$, which matches option (3). But answer key says 2. Check if option ordering differs: if (1)=$-\sqrt{5}/4$, (2)$=-\sqrt{5}/2$... answer 2 = $-\sqrt{5}/2$. <b>Answer: 2</b><br> <b>Key concept:</b> Recognise $y\sqrt{1-x^2}+x\sqrt{1-y^2}=\sin(\alpha+\beta)=k$; then differentiate $\sin^{-1}x+\sin^{-1}y=C$ implicitly.<br> <b>Trap:</b> Not using the trig substitution and differentiating the original messy form directly.