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Ответ:

Производная сложной функции :   [tex]\bf (ln\, u)'=\dfrac{1}{u}\cdot u'[/tex]  ,   [tex]\bf (tgu)'=\dfrac{1}{cos^2u}\cdot u'[/tex] ,

    [tex]\bf (e^{u})'=e^{u}\cdot u'[/tex]   .  

[tex]\bf y=ln\Big(\dfrac{1}{x}+tg\, e^{sinx}\Big)\\\\\\y'=\dfrac{1}{\dfrac{1}{x}+tg\, e^{sinx}}\cdot \Big(\dfrac{1}{x}+tg\, e^{sinx}\Big)'=\\\\\\=\dfrac{x}{1+x\cdot tg\, e^{sinx}}\cdot \Big(-\dfrac{1}{x^2}+\dfrac{1}{cos^2(e^{sinx})}\cdot (e^{sinx})'\Big)=\\\\\\=\dfrac{x}{1+x\cdot tg\, e^{sinx}}\cdot \Big(-\dfrac{1}{x^2}+\dfrac{1}{cos^2(e^{sinx})}\cdot e^{sinx}\cdot (sinx)'\Big)=[/tex]  

[tex]\bf =\dfrac{x}{1+x\cdot tg\, e^{sinx}}\cdot \Big(-\dfrac{1}{x^2}+\dfrac{1}{cos^2(e^{sinx})}\cdot e^{sinx}\cdot cosx\Big)[/tex]