Differentiation: ACJC 2025 Promo Q1

Taking logarithms of $y^{\cos x} = x^{\sin 2x}$: $$\begin{aligned} (\cos x)\ln y &= (\sin 2x)\ln x. \end{aligned}$$ Differentiating both sides with respect to $x$: $$\begin{aligned} (-\sin x)\ln y + (\cos x)\frac{1}{y}\frac{\,\mathrm{d} y}{\,\mathrm{d} x} &= (2\cos 2x)\ln x + \frac{\sin 2x}{x}. \quad\text{(1)} \end{aligned}$$ At $x = \pi$: $y^{\cos\pi} = \pi^{\sin 2\pi}$, so $y^{-1} = 1$ and $y = 1$.

Substituting $x = \pi$, $y = 1$ into (1): $$\begin{aligned} 0 + (\cos\pi)(1)\frac{\,\mathrm{d} y}{\,\mathrm{d} x} &= (2\cos 2\pi)\ln\pi + 0\\ -\frac{\,\mathrm{d} y}{\,\mathrm{d} x} &= 2\ln\pi\\ \frac{\,\mathrm{d} y}{\,\mathrm{d} x} &= -2\ln\pi. \end{aligned}$$

As the base is not a constant, do not apply $\dfrac{\,\mathrm{d}}{\,\mathrm{d} x}a^{\mathrm{f}(x)} = \mathrm{f}'(x)\,a^{\mathrm{f}(x)}\ln a$. Leave the answer in exact form, not to 3 s.f.