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Question Number 110456 by bobhans last updated on 29/Aug/20

If (f(x).g(x))′ = f(x)′ . g(x)′ find the function of f(x) .

$$\:\:\:{If}\:\left({f}\left({x}\right).{g}\left({x}\right)\right)'\:=\:{f}\left({x}\right)'\:.\:{g}\left({x}\right)'\: \\ $$$$\:{find}\:{the}\:{function}\:{of}\:{f}\left({x}\right)\:. \\ $$

Answered by bemath last updated on 29/Aug/20

(d/dx)(f(x).g(x))=f(x)′.g(x)+f(x).g(x)′ then given condition ⇒f(x)′.g(x)+f(x).g′(x)=f(x)′.g(x)′ ⇒f(x)′.g(x)−f(x)′.g(x)′=−f(x).g(x)′ ⇒f(x)′(g(x)−g(x)′)=−f(x).g(x)′ ⇒ ((f(x)′)/(f(x))) = ((g(x)′)/(g(x)′−g(x))) ∫ ((f(x)′)/(f(x)))dx = ∫ ((g(x)′)/(g(x)′−g(x)))dx ⇒ln (f(x))= ∫((g(x)′)/(g(x)′−g(x)))dx+c ⇒f(x) = C.exp(∫((g(x)′)/(g(x)′−g(x)))dx) ∴ f(x) = C.e^(∫(((g(x)′)/(g(x)′−g(x)))dx))

$$\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{f}\left(\mathrm{x}\right).\mathrm{g}\left(\mathrm{x}\right)\right)=\mathrm{f}\left(\mathrm{x}\right)'.\mathrm{g}\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{x}\right).\mathrm{g}\left(\mathrm{x}\right)' \\ $$$$\mathrm{then}\:\mathrm{given}\:\mathrm{condition}\: \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{x}\right)'.\mathrm{g}\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{x}\right).\mathrm{g}'\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{x}\right)'.\mathrm{g}\left(\mathrm{x}\right)' \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{x}\right)'.\mathrm{g}\left(\mathrm{x}\right)−\mathrm{f}\left(\mathrm{x}\right)'.\mathrm{g}\left(\mathrm{x}\right)'=−\mathrm{f}\left(\mathrm{x}\right).\mathrm{g}\left(\mathrm{x}\right)' \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{x}\right)'\left(\mathrm{g}\left(\mathrm{x}\right)−\mathrm{g}\left(\mathrm{x}\right)'\right)=−\mathrm{f}\left(\mathrm{x}\right).\mathrm{g}\left(\mathrm{x}\right)' \\ $$$$\Rightarrow\:\frac{\mathrm{f}\left(\mathrm{x}\right)'}{\mathrm{f}\left(\mathrm{x}\right)}\:=\:\frac{\mathrm{g}\left(\mathrm{x}\right)'}{\mathrm{g}\left(\mathrm{x}\right)'−\mathrm{g}\left(\mathrm{x}\right)} \\ $$$$\int\:\frac{\mathrm{f}\left(\mathrm{x}\right)'}{\mathrm{f}\left(\mathrm{x}\right)}\mathrm{dx}\:=\:\int\:\frac{\mathrm{g}\left(\mathrm{x}\right)'}{\mathrm{g}\left(\mathrm{x}\right)'−\mathrm{g}\left(\mathrm{x}\right)}\mathrm{dx} \\ $$$$\Rightarrow\mathrm{ln}\:\left(\mathrm{f}\left(\mathrm{x}\right)\right)=\:\int\frac{\mathrm{g}\left(\mathrm{x}\right)'}{\mathrm{g}\left(\mathrm{x}\right)'−\mathrm{g}\left(\mathrm{x}\right)}\mathrm{dx}+\mathrm{c} \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{C}.\mathrm{exp}\left(\int\frac{\mathrm{g}\left(\mathrm{x}\right)'}{\mathrm{g}\left(\mathrm{x}\right)'−\mathrm{g}\left(\mathrm{x}\right)}\mathrm{dx}\right) \\ $$$$\therefore\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{C}.\mathrm{e}^{\int\left(\frac{\mathrm{g}\left(\mathrm{x}\right)'}{\mathrm{g}\left(\mathrm{x}\right)'−\mathrm{g}\left(\mathrm{x}\right)}\mathrm{dx}\right)} \\ $$

Commented by bobhans last updated on 29/Aug/20

great....

$$\mathrm{great}.... \\ $$

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