Question Number 69680 by 20190927 last updated on 26/Sep/19
$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{sinx}} \:\:,\:\mathrm{0}<\mathrm{x}<\frac{\pi}{\mathrm{2}}\:\:\:\mathrm{find}\:\mathrm{f}'\left(\mathrm{x}\right) \\ $$
Answered by MJS last updated on 26/Sep/19
![(d/dx)[u^v ]=u^v (((u′v)/u)+v′ln u) ⇒ f′(x)=x^(sin x) (((sin x)/x)+cos x ln x)](https://www.tinkutara.com/question/Q69683.png)
$$\frac{{d}}{{dx}}\left[{u}^{{v}} \right]={u}^{{v}} \left(\frac{{u}'{v}}{{u}}+{v}'\mathrm{ln}\:{u}\right) \\ $$$$\Rightarrow \\ $$$${f}'\left({x}\right)={x}^{\mathrm{sin}\:{x}} \left(\frac{\mathrm{sin}\:{x}}{{x}}+\mathrm{cos}\:{x}\:\mathrm{ln}\:{x}\right) \\ $$
Commented by 20190927 last updated on 26/Sep/19
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$
Answered by Henri Boucatchou last updated on 26/Sep/19
$$\boldsymbol{{f}}'\left(\boldsymbol{{x}}\right)\:=\:\left(\boldsymbol{{e}}^{\boldsymbol{{sinxlnx}}} \right)'=\:\left(\boldsymbol{{sinxlnx}}\right)'\boldsymbol{{f}}\left(\boldsymbol{{x}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\:\left(\boldsymbol{{cosxlnx}}+\frac{\boldsymbol{{sinx}}}{\boldsymbol{{x}}}\right)\boldsymbol{{x}}^{\boldsymbol{{sinx}}} \\ $$