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Derivative of (2-2x)exp^(2x-x^2)

Function f() - derivative -N order at the point
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The solution

You have entered [src]
                  2
           2*x - x 
(2 - 2*x)*E        
$$e^{- x^{2} + 2 x} \left(2 - 2 x\right)$$
(2 - 2*x)*E^(2*x - x^2)
Detail solution
  1. Apply the product rule:

    ; to find :

    1. Differentiate term by term:

      1. The derivative of the constant is zero.

      2. The derivative of a constant times a function is the constant times the derivative of the function.

        1. Apply the power rule: goes to

        So, the result is:

      The result is:

    ; to find :

    1. Let .

    2. The derivative of is itself.

    3. Then, apply the chain rule. Multiply by :

      1. Differentiate term by term:

        1. The derivative of a constant times a function is the constant times the derivative of the function.

          1. Apply the power rule: goes to

          So, the result is:

        2. The derivative of a constant times a function is the constant times the derivative of the function.

          1. Apply the power rule: goes to

          So, the result is:

        The result is:

      The result of the chain rule is:

    The result is:

  2. Now simplify:


The answer is:

The graph
The first derivative [src]
            2                      2
     2*x - x             2  2*x - x 
- 2*e         + (2 - 2*x) *e        
$$\left(2 - 2 x\right)^{2} e^{- x^{2} + 2 x} - 2 e^{- x^{2} + 2 x}$$
The second derivative [src]
           /   x*(2 - x)   /               2\  -x*(-2 + x)\
4*(-1 + x)*\2*e          - \-1 + 2*(-1 + x) /*e           /
$$4 \left(x - 1\right) \left(- \left(2 \left(x - 1\right)^{2} - 1\right) e^{- x \left(x - 2\right)} + 2 e^{x \left(2 - x\right)}\right)$$
The third derivative [src]
  /              2             2 /               2\\  -x*(-2 + x)
4*\3 - 6*(-1 + x)  + 2*(-1 + x) *\-3 + 2*(-1 + x) //*e           
$$4 \left(2 \left(x - 1\right)^{2} \left(2 \left(x - 1\right)^{2} - 3\right) - 6 \left(x - 1\right)^{2} + 3\right) e^{- x \left(x - 2\right)}$$