Differentiation: TJC 2024 Promo Q5

(ai)

$$\begin{aligned} \frac{\,\mathrm{d}}{\,\mathrm{d} x}\ln\!\left(\frac{1+\sin x}{1-\sin x}\right) &=\frac{\,\mathrm{d}}{\,\mathrm{d} x}\bigl(\ln(1+\sin x)-\ln(1-\sin x)\bigr)\\ &=\frac{\cos x}{1+\sin x}-\frac{-\cos x}{1-\sin x}\\ &=\frac{\cos x(1-\sin x)+\cos x(1+\sin x)}{(1+\sin x)(1-\sin x)}\\ &=\frac{2\cos x}{1-\sin^2x}\\ &=\frac{2\cos x}{\cos^2x}\\ &=2\sec x \end{aligned}$$

Q5(aii)

$$\begin{aligned} \frac{\,\mathrm{d}}{\,\mathrm{d} x}\tan^{-1}\!\left(\frac{1}{\sqrt{1-x}}\right) &=\frac{1}{1+\left(\frac{1}{\sqrt{1-x}}\right)^2} \frac{\,\mathrm{d}}{\,\mathrm{d} x}\left(\frac{1}{\sqrt{1-x}}\right)\\ &=\frac{1-x}{(1-x)+1}\left(-\frac{1}{2(1-x)^{3/2}}(-1)\right)\\ &=\frac{1}{2(2-x)\sqrt{1-x}} \end{aligned}$$

5(b)

$y=x^{3x}$.

Taking logarithm on both sides: $$\begin{aligned} \ln y&=\ln x^{3x},\\ \ln y&=3x\ln x. \end{aligned}$$ Differentiating implicitly w.r.t. $x$ on both sides: $$\begin{aligned} \frac1y\frac{\,\mathrm{d} y}{\,\mathrm{d} x}&=(3)\ln x+3x\left(\frac1x\right)\\ \frac1y\frac{\,\mathrm{d} y}{\,\mathrm{d} x}&=3\ln x+3,\\ \frac{\,\mathrm{d} y}{\,\mathrm{d} x}&=y(3\ln x+3),\\ &=3x^{3x}(\ln x+1). \end{aligned}$$