Question Number 121720 by Jamshidbek2311 last updated on 11/Nov/20
$${Does}\:{this}\:{example}\:{work}\:{that}\:{way}. \\ $$$${f}\left({x}\right)={x}^{{x}} \:\:\:{f}'\left({x}\right)={x}^{{x}} \centerdot\left({lnx}+\mathrm{1}\right) \\ $$
Answered by Dwaipayan Shikari last updated on 11/Nov/20
$${f}\left({x}\right)={x}^{{x}} \\ $$$${log}\left({f}\left({x}\right)\right)={xlogx} \\ $$$$\frac{{f}'\left({x}\right)}{{f}\left({x}\right)}=\mathrm{1}+{logx} \\ $$$${f}'\left({x}\right)={f}\left({x}\right)\left(\mathrm{1}+{logx}\right)={x}^{{x}} \left(\mathrm{1}+{logx}\right) \\ $$
Answered by ebi last updated on 11/Nov/20
$$ \\ $$$${Yes}. \\ $$$${f}\left({x}\right)={x}^{{x}} \\ $$$${let}\:{y}={x}^{{x}} \\ $$$${ln}\:{y}={ln}\:{x}^{{x}} \\ $$$${ln}\:{y}={x}\:{ln}\:{x} \\ $$$$\frac{{d}}{{dx}}\left({ln}\:{y}\right)=\frac{{d}}{{dx}}\left({x}\:{ln}\:{x}\right) \\ $$$$\frac{\mathrm{1}}{{y}}\centerdot\frac{{dy}}{{dx}}={x}\centerdot\frac{\mathrm{1}}{{x}}+{ln}\:{x} \\ $$$$\frac{\mathrm{1}}{{y}}\centerdot\frac{{dy}}{{dx}}=\mathrm{1}+{ln}\:{x} \\ $$$$\frac{{dy}}{{dx}}={y}\left(\mathrm{1}+{ln}\:{x}\right) \\ $$$$\therefore{f}'\left({x}\right)={x}^{{x}} \left(\mathrm{1}+{ln}\:{x}\right) \\ $$
Commented by Jamshidbek2311 last updated on 11/Nov/20
$${thank} \\ $$