Using the differentiation rules and the derivative table, find the derivative of the function y=(x-1)e^{-\frac{1}{x}}


y'=e^{-\frac{1}{x}} \left( \frac{x^2 + x - 1}{x^2} \right)



This is the product of two functions f\cdot g, with f(x)= x-1 and g(x)= e^{-\frac{1}{x}}

  • Differentiate the factors

    - Derivative of f(x)=(x-1): f'(x)= 1

    - Derivative of g(x)=e^{-\frac{1}{x}}

    This is a composition of functions, of the type g(x)=e^{v(x)}, with inner function v(x)=\frac{1}{x}. Observing that v'(x)=\frac{d}{dx}x^{-1}=-1\cdot x^{-2}, from the Chain Rule we have

    g'(x)=e^{-\frac{1}{x}} \cdot \frac{1}{x^2}


  • Apply the Product Rule

    The Product Rule is [f\cdot g]'=f'g+fg'. Substituting f e g and their derivatives we find

    y' = 1 \cdot e^{-\frac{1}{x}} + (x-1) \cdot \frac{e^{-\frac{1}{x}}}{x^2}


  • Simplify the result

    \begin{aligned} y' &= e^{-\frac{1}{x}} + \frac{x-1}{x^2} e^{-\frac{1}{x}}\\ & = e^{-\frac{1}{x}} \left( 1 + \frac{x-1}{x^2} \right)\\ & = e^{-\frac{1}{x}} \left( \frac{x^2 + x - 1}{x^2} \right)\qquad \text{final result} \end{aligned}