Answer:
[tex]\huge\boxed{f'(x)=-\dfrac{e^\frac{1}{x}}{x^2}}[/tex]
Step-by-step explanation:
[tex](e^x)'=e^x\\\\\bigg[f\bigg(g(x)\bigg)\bigg]'=f'\bigg(g(x)\bigg)\cdot g'(x)\\\\\left(\dfrac{1}{x}\right)'=-\dfrac{1}{x^2}\\\\====================[/tex]
[tex]f(x)=e^\frac{1}{x}=\left(e^\frac{1}{x}\right)'\cdot\left(\dfrac{1}{x}\right)'=e^\frac{1}{x}\cdot\left(-\dfrac{1}{x^2}\right)=-\dfrac{e^\frac{1}{x}}{x^2}[/tex]
