It is given that $y=\ln \left( 1+{{\mathrm{e}}^{x}} \right)$. Show that \(\left( 1+{{\ma

Question

It is given that $y=\ln \left( 1+{{\mathrm{e}}^{x}} \right)$. Show that $\left( 1+{{\mathrm{e}}^{x}} \right)\frac{\mathrm{d}y}{\mathrm{d}x}-{{\mathrm{e}}^{x}}=0$.[1] By further differentiation of the result in (i), find the Maclaurin series for $y$, up to and including the term in ${{x}^{2}}$. Hence find the series for $\frac{{{\mathrm{e}}^{x}}}{1+{{\mathrm{e}}^{x}}}$ up to and including $x$.[4] Using appropriate expansion from the List of Formulae (MF26), verify the correctness of the series $\frac{{{\mathrm{e}}^{x}}}{1+{{\mathrm{e}}^{x}}}$ found in (ii).[3]

Accepted Answer

(ii) $\ln(1+\mathrm e^x)=\ln2+\dfrac12x+\dfrac18x^2+\cdots$, hence $\dfrac{\mathrm e^x}{1+\mathrm e^x}=\dfrac12+\dfrac14x+\cdots$

2021 NYJC P1 Q7

2021 NYJC P1 Q7

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