Question Number 2071 by Rasheed Soomro last updated on 01/Nov/15
![{ [f(x)]^(−1) }′ =[ f ′(x) ]^(−1) f(x)=?](https://www.tinkutara.com/question/Q2071.png)
$$\left\{\:\left[{f}\left({x}\right)\right]^{−\mathrm{1}} \right\}'\:=\left[\:{f}\:'\left({x}\right)\:\right]^{−\mathrm{1}} \\ $$$${f}\left({x}\right)=? \\ $$
Commented by Yozzi last updated on 01/Nov/15
$$\frac{{d}}{{dx}}\left(\frac{\mathrm{1}}{{f}\left({x}\right)}\right)=\frac{\mathrm{1}}{{f}^{'} \left({x}\right)} \\ $$$$\frac{{f}\left({x}\right)×\mathrm{0}−{f}^{'} \left({x}\right)}{\left({f}\left({x}\right)\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{{f}^{'} \left({x}\right)} \\ $$$$−\left({f}^{'} \left({x}\right)\right)^{\mathrm{2}} =\left({f}\left({x}\right)\right)^{\mathrm{2}} \\ $$$$\left({f}\left({x}\right)\right)^{\mathrm{2}} +\left({f}^{'} \left({x}\right)\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${f}\left({x}\right)=\pm{if}^{'} \left({x}\right) \\ $$$$\mathrm{1}=\pm\frac{{if}^{'} \left({x}\right)}{{f}\left({x}\right)} \\ $$$$\int\mathrm{1}{dx}=\pm{i}\int\frac{{f}^{'} \left({x}\right)}{{f}\left({x}\right)}{dx} \\ $$$$\pm{iln}\mid{f}\left({x}\right)\mid={x}+{c} \\ $$$${ln}\mid{f}\left({x}\right)\mid=\mp\left({i}\left({x}+{c}\right)\right) \\ $$$${f}\left({x}\right)={e}^{\mp{i}\left({x}+{c}\right)} \\ $$$${f}\left({x}\right)={cos}\left({x}+{c}\right)+{isin}\left({x}+{c}\right) \\ $$$${or} \\ $$$${f}\left({x}\right)={cos}\left({x}+{c}\right)−{isin}\left({x}+{c}\right) \\ $$$${f}^{'} \left({x}\right)={ie}^{{i}\left({x}+{c}\right)} \Rightarrow\frac{\mathrm{1}}{{f}^{'} \left({x}\right)}=−{ie}^{−{i}\left({x}+{c}\right)} \\ $$$$\frac{\mathrm{1}}{{f}\left({x}\right)}={e}^{−{i}\left({x}+{c}\right)} \Rightarrow\frac{{d}}{{dx}}\left(\frac{\mathrm{1}}{{f}\left({x}\right)}\right)=−{ie}^{−{i}\left({x}+{c}\right.} ={f}^{'} \left({x}\right) \\ $$$$ \\ $$
Commented by Rasheed Soomro last updated on 01/Nov/15
$$\mathcal{L}{ike}\:{your}\:{approach}! \\ $$