It is given that $y = x^{xy}$, where $x 0$, $y 0$. Show that \(\frac{

Question

It is given that $y = x^{xy}$, where $x > 0$, $y > 0$. Show that $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{y^2 (1 + \ln x)}{1 - \ln y}$.[3] Hence find the coordinates of the point on the curve $y = x^{xy}$ whose tangent is parallel to the $y$-axis.[2]

Accepted Answer

(a) $\displaystyle\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{y^2(1+\ln x)}{1-\ln y}$ (b) $(1.32,2.72)$ (3 s.f.)

2025 ACJC P1 Q3

2025 ACJC P1 Q3

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