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Question Number 2071 by Rasheed Soomro last updated on 01/Nov/15

{ [f(x)]^(−1) }′ =[ f ′(x) ]^(−1) f(x)=?

$$\left\{\:\left[{f}\left({x}\right)\right]^{−\mathrm{1}} \right\}'\:=\left[\:{f}\:'\left({x}\right)\:\right]^{−\mathrm{1}} \\ $$$${f}\left({x}\right)=? \\ $$

Commented by Yozzi last updated on 01/Nov/15

(d/dx)((1/(f(x))))=(1/(f^′ (x))) ((f(x)×0−f^′ (x))/((f(x))^2 ))=(1/(f^′ (x))) −(f^′ (x))^2 =(f(x))^2 (f(x))^2 +(f^′ (x))^2 =0 f(x)=±if^′ (x) 1=±((if^′ (x))/(f(x))) ∫1dx=±i∫((f^′ (x))/(f(x)))dx ±iln∣f(x)∣=x+c ln∣f(x)∣=∓(i(x+c)) f(x)=e^(∓i(x+c)) f(x)=cos(x+c)+isin(x+c) or f(x)=cos(x+c)−isin(x+c) f^′ (x)=ie^(i(x+c)) ⇒(1/(f^′ (x)))=−ie^(−i(x+c)) (1/(f(x)))=e^(−i(x+c)) ⇒(d/dx)((1/(f(x))))=−ie^(−i(x+c) =f^′ (x)

$$\frac{{d}}{{dx}}\left(\frac{\mathrm{1}}{{f}\left({x}\right)}\right)=\frac{\mathrm{1}}{{f}^{'} \left({x}\right)} \\ $$$$\frac{{f}\left({x}\right)×\mathrm{0}−{f}^{'} \left({x}\right)}{\left({f}\left({x}\right)\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{{f}^{'} \left({x}\right)} \\ $$$$−\left({f}^{'} \left({x}\right)\right)^{\mathrm{2}} =\left({f}\left({x}\right)\right)^{\mathrm{2}} \\ $$$$\left({f}\left({x}\right)\right)^{\mathrm{2}} +\left({f}^{'} \left({x}\right)\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${f}\left({x}\right)=\pm{if}^{'} \left({x}\right) \\ $$$$\mathrm{1}=\pm\frac{{if}^{'} \left({x}\right)}{{f}\left({x}\right)} \\ $$$$\int\mathrm{1}{dx}=\pm{i}\int\frac{{f}^{'} \left({x}\right)}{{f}\left({x}\right)}{dx} \\ $$$$\pm{iln}\mid{f}\left({x}\right)\mid={x}+{c} \\ $$$${ln}\mid{f}\left({x}\right)\mid=\mp\left({i}\left({x}+{c}\right)\right) \\ $$$${f}\left({x}\right)={e}^{\mp{i}\left({x}+{c}\right)} \\ $$$${f}\left({x}\right)={cos}\left({x}+{c}\right)+{isin}\left({x}+{c}\right) \\ $$$${or} \\ $$$${f}\left({x}\right)={cos}\left({x}+{c}\right)−{isin}\left({x}+{c}\right) \\ $$$${f}^{'} \left({x}\right)={ie}^{{i}\left({x}+{c}\right)} \Rightarrow\frac{\mathrm{1}}{{f}^{'} \left({x}\right)}=−{ie}^{−{i}\left({x}+{c}\right)} \\ $$$$\frac{\mathrm{1}}{{f}\left({x}\right)}={e}^{−{i}\left({x}+{c}\right)} \Rightarrow\frac{{d}}{{dx}}\left(\frac{\mathrm{1}}{{f}\left({x}\right)}\right)=−{ie}^{−{i}\left({x}+{c}\right.} ={f}^{'} \left({x}\right) \\ $$$$ \\ $$

Commented by Rasheed Soomro last updated on 01/Nov/15

Like your approach!

$$\mathcal{L}{ike}\:{your}\:{approach}! \\ $$

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