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Question Number 172190 by Mikenice last updated on 23/Jun/22

Commented by mokys last updated on 23/Jun/22

$$ln\left(f\left(x\right)\right)=cos(x^2+4x+2)lnx$$$$$$$$\frac{f{}^{{}^{\prime }}\left(x\right)}{f\left(x\right)}=\frac{cos(x^2+4x+2)}{x}-(2x+4)sin(x^2+4x+2)lnx$$$$$$$$f{}^{{}^{\prime }}\left(x\right)=f\left(x\right)\left[\frac{cos(x^2+4x+2)-(2x^2+4x)lnxsin(x^2+4x+2)}{x}\right]$$$$$$$$f{}^{{}^{\prime }}\left(x\right)=X{}^{cos(x^2+4x+2)}\left[\frac{cos(x^2+4x+2)-(2x^2+4x)lnxsin(x^2+4x+2)}{x}\right]$$$$$$$$Aldolimy$$

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