$\frac{\text{x}+1}{\text{x}+2}$
$\frac{\text{x}+1}{\text{x}+2}$
$\text{f(x)}=\frac{\text{x}+1}{\text{x}+2}$
$\because\text{f}'\text{(x)}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f(x+h)}-\text{f(x)}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\frac{\text{(x+h)+1}}{\text{(x+h)+2}}-\frac{\text{x}+1}{\text{x}+2}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{(x+2)}\text{(x+h+1)}-\text{(x+1)(x+h+2)}}{\text{(x+h+2)}\text{(x+2)h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\big(\text{x}^2+2\text{x+xh+2h+2+x}\big)-\big(\text{x}^2+\text{xh+2x+x+h+2}\big)}{\text{(x+h+2)(x+2)h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}}{\text{(x+h+2)(x+2)h}}$
$=\frac{1}{(\text{x}+2)^2}$
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$\frac{1+3^\text{x}}{1-3^\text{x}}$
$9\text{x}^{2}-16\text{y}^{2}=-144$
If the term from x in the expansion of $\Big(\sqrt{\text{x}}-\frac{\text{k}}{\text{x}^{2}}\Big)^{10}$ is 405, find the value of k.