这是一份kcl伦敦大学学院 4CCM111A作业代写的成功案
so
$e^{y}-2 x-e^{-y}=0$
or, multiplying by $e^{y}$,
$$
e^{2 y}-2 x e^{y}-1=0
$$
This is really a quadratic equation in $e^{y}$ :
$$
\left(e^{y}\right)^{2}-2 x\left(e^{y}\right)-1=0
$$
Solving by the quadratic formula, we get
$$
e^{y}=\frac{2 x \pm \sqrt{4 x^{2}+4}}{2}=x \pm \sqrt{x^{2}+1}
$$
Note that $e^{y}>0$, but $x-\sqrt{x^{2}+1}<0$ (because $x<\sqrt{x^{2}+1}$ ). Thus the minus sign is inadmissible and we have
$$
e^{y}=x+\sqrt{x^{2}+1}
$$
Therefore
$$
y=\ln \left(e^{y}\right)=\ln \left(x+\sqrt{x^{2}+1}\right)
$$
4CCM111A COURSE NOTES ：
Using Table 6 and the Chain Rule, we have
$$
\begin{aligned}
\frac{d}{d x}\left[\tanh ^{-1}(\sin x)\right] &=\frac{1}{1-(\sin x)^{2}} \frac{d}{d x}(\sin x) \
&=\frac{1}{1-\sin ^{2} x} \cos x=\frac{\cos x}{\cos ^{2} x}=\sec x
\end{aligned}
$$