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Worked solution

Implicit differentiation involving inverse sine

A curve has equation sin⁡−1y+xey=3x\sin^{-1}y+xe^y=3x. Find dydx\dfrac{dy}{dx} in terms of xx and yy.

[2009/CJC/Promo/10(b)]

CJC · Discussion · Question 4

Answer

dydx=(3−ey)1−y21+xey1−y2\dfrac{dy}{dx}=\dfrac{(3-e^y)\sqrt{1-y^2}}{1+xe^y\sqrt{1-y^2}}.

Differentiate and collect

For ∣y∣<1|y|<1, differentiate to obtain 11−y2dydx+xeydydx+ey=3\dfrac1{\sqrt{1-y^2}}\dfrac{dy}{dx}+xe^y\dfrac{dy}{dx}+e^y=3. Multiply through by 1−y2\sqrt{1-y^2}: dydx+xey1−y2dydx=(3−ey)1−y2\dfrac{dy}{dx}+xe^y\sqrt{1-y^2}\dfrac{dy}{dx}=(3-e^y)\sqrt{1-y^2}. Thus dydx=(3−ey)1−y21+xey1−y2\dfrac{dy}{dx}=\dfrac{(3-e^y)\sqrt{1-y^2}}{1+xe^y\sqrt{1-y^2}} where the denominator is nonzero.

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Common mistakes

The displayed derivation uses the differentiable interior |y|<1; do not infer endpoint behaviour just by substituting into the simplified formula.

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