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f-x-ln-x-x-x-2-g-x-x-2-1-ln-x-Calculate-the-derivative-of-f-x-as-a-function-of-g-x-




Question Number 171546 by Kodjo last updated on 17/Jun/22
f(x)=((−ln∣x∣)/x)+x−2  ,   g(x)=−x^2 +1−ln∣x∣    Calculate the derivative of f(x) as a function of g(x)
$${f}\left({x}\right)=\frac{−{ln}\mid{x}\mid}{{x}}+{x}−\mathrm{2}\:\:,\:\:\:{g}\left({x}\right)=−{x}^{\mathrm{2}} +\mathrm{1}−{ln}\mid{x}\mid \\ $$$$ \\ $$Calculate the derivative of f(x) as a function of g(x)
Commented by kaivan.ahmadi last updated on 17/Jun/22
f(x)=((−lnx+x^2 −2x)/x)=((g(x)+x^2 −1+x^2 −2x)/x)  =((g(x)−2x−1)/x)=((g(x))/x)−2−(1/x)  f′(x)=((xg′(x)−g(x))/x^2 )+(1/x^2 )=((xg′(x)−g(x)+1)/x^2 )
$${f}\left({x}\right)=\frac{−{lnx}+{x}^{\mathrm{2}} −\mathrm{2}{x}}{{x}}=\frac{{g}\left({x}\right)+{x}^{\mathrm{2}} −\mathrm{1}+{x}^{\mathrm{2}} −\mathrm{2}{x}}{{x}} \\ $$$$=\frac{{g}\left({x}\right)−\mathrm{2}{x}−\mathrm{1}}{{x}}=\frac{{g}\left({x}\right)}{{x}}−\mathrm{2}−\frac{\mathrm{1}}{{x}} \\ $$$${f}'\left({x}\right)=\frac{{xg}'\left({x}\right)−{g}\left({x}\right)}{{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\frac{{xg}'\left({x}\right)−{g}\left({x}\right)+\mathrm{1}}{{x}^{\mathrm{2}} } \\ $$

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