Using the differentiation rules and the derivative table, find the derivative of the function y=\displaystyle\frac{e^{-x^2}}{2-x}


y' = \frac{e^{-x^2} (1 - 4x + 2x^2)}{(2 - x)^2}

We can use the Quotient Rule. The quotient rule states that if y = \frac{f(x)}{g(x)} with f and g differentiable, then the derivative y' is given by:

y' = \frac{f'(x)g(x) - f(x)g'(x)}{g^2(x)}

  • Identify the functions

    f(x) = e^{-x^2}, \quad g(x) = 2 - x


  • Calculate the derivatives of f and g

    - Derivative of f: using the Chain Rule (we write f as f(x)=e^{v(x)} with inner function v(x)=-x^2)

    f'(x) = e^{-x^2} \cdot \frac{d}{dx}(-x^2) = e^{-x^2} \cdot (-2x) = -2x e^{-x^2}


    - Derivative of g: the function is g(x) = 2 - x so its derivative is

    g'(x) = -1


  • Apply the Quotient Rule

    Substituting f, g, f' and g' in the formula we find

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

  • Simplify

    \begin{aligned} y' &= \frac{-2x e^{-x^2} (2 - x) + e^{-x^2}}{(2 - x)^2}\\ &= \frac{-4x e^{-x^2} + 2x^2 e^{-x^2} + e^{-x^2}}{(2 - x)^2}\\ & = \frac{e^{-x^2} (1 - 4x + 2x^2)}{(2 - x)^2}\qquad \text{final result} \end{aligned}