Question Number 3764 by 123456 last updated on 19/Dec/15
$$\mathrm{if}\:\frac{{df}}{{dx}}={f},\:\mathrm{what}\:\mathrm{are}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\frac{{d}\theta}{{dx}}?\:\left(\mathrm{Q3707}\right) \\ $$
Commented by Filup last updated on 19/Dec/15
$$\mathrm{Attempt}: \\ $$$$ \\ $$$$\frac{{d}\theta}{{dx}}=\frac{\frac{{d}^{\mathrm{2}} {f}}{{dx}^{\mathrm{2}} }}{\mathrm{1}+\left(\frac{{df}}{{dx}}\right)^{\mathrm{2}} } \\ $$$$\left(\frac{{df}}{{dx}}\right)^{\mathrm{2}} ={f}^{\mathrm{2}} \\ $$$$\frac{{d}^{\mathrm{2}} {f}}{{dx}^{\mathrm{2}} }={f}' \\ $$$$ \\ $$$$\therefore\frac{{d}\theta}{{dx}}=\frac{{f}'}{\mathrm{1}+{f}^{\mathrm{2}} } \\ $$
Commented by prakash jain last updated on 20/Dec/15
$$\frac{{df}}{{dx}}={f} \\ $$$${f}\left({x}\right)={e}^{{x}+{c}} \\ $$$$\mathrm{tan}\:\theta={f}\:'\left({x}\right)={e}^{{x}+{c}} \\ $$$$\frac{\mathrm{d}\theta}{\mathrm{d}{x}}=\frac{{e}^{{x}+{c}} }{\mathrm{1}+{e}^{\mathrm{2}{x}+\mathrm{2}{c}} } \\ $$