Given that $y=\frac{1}{\sqrt{2{{\mathrm{e}}^{x}}-1}}$, show that $\frac{\mathrm{d}y}{\mathrm{d}x}=-{

Question

Given that $y=\frac{1}{\sqrt{2{{\mathrm{e}}^{x}}-1}}$, show that $\frac{\mathrm{d}y}{\mathrm{d}x}=-{{y}^{3}}{{\mathrm{e}}^{x}}$, [2] $y\frac{{{\mathrm{d}}^{2}}y}{\mathrm{d}{{x}^{2}}}=y\frac{\mathrm{d}y}{\mathrm{d}x}+3{{\left( \frac{\mathrm{d}y}{\mathrm{d}x} \right)}^{2}}$. [2] Find $\frac{\mathrm{d}}{\mathrm{d}x}\left( \ln \left( \frac{2x}{\sqrt{{{x}^{2}}+1}} \right) \right)$. [2]

Accepted Answer

$\frac{dy}{dx} = -y^3 e^x$; $y\frac{d^2y}{dx^2} = y\frac{dy}{dx} + 3\left(\frac{dy}{dx}\right)^2$; $\frac{1}{x} - \frac{x}{x^2+1}$

2022 SAJC Promo Q3

2022 SAJC Promo Q3

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