How do you find the 1000th derivative of #y=xe^-x#?

Try to determine a general pattern for the derivative of the function by taking the first few derivatives.

So, we have a prety good pattern going.

To find the 1000th derivative of ( y = xe^{-x} ), you can use the formula for the nth derivative of a function multiplied by a polynomial. In this case, the function is ( e^{-x} ).

The nth derivative of ( e^{-x} ) is ( (-1)^n e^{-x} ), where ( n ) is a non-negative integer.

Now, for the function ( xe^{-x} ), the nth derivative can be found using the product rule, which states that if you have two functions ( u ) and ( v ), the nth derivative of their product is given by:

[ (uv)^{(n)} = \sum_{k=0}^{n} \binom{n}{k} u^{(n-k)}v^{(k)} ]

where ( u^{(n-k)} ) denotes the nth derivative of ( u ) and ( v^{(k)} ) denotes the kth derivative of ( v ).

In this case, ( u = x ) and ( v = e^{-x} ).

So, applying the product rule to find the 1000th derivative, we have:

[ y^{(1000)} = \sum_{k=0}^{1000} \binom{1000}{k} (x^{(1000-k)})(e^{-x})^{(k)} ]

Now, for ( x^{(1000-k)} ), it's a constant term if ( 1000 - k = 0 ), meaning ( k = 1000 ).

[ x^{(1000-k)} = x^{(0)} = x ]

For ( (e^{-x})^{(k)} ), we know that the kth derivative of ( e^{-x} ) is ( (-1)^k e^{-x} ).

[ (e^{-x})^{(k)} = (-1)^k e^{-x} ]

So, substituting these into the formula, we get:

[ y^{(1000)} = \binom{1000}{1000} (x)(-1)^{1000} e^{-x} = x(-1)^{1000} e^{-x} ]

[ y^{(1000)} = (-1)^{1000} xe^{-x} ]

[ y^{(1000)} = xe^{-x} ]

Therefore, the 1000th derivative of ( y = xe^{-x} ) is ( xe^{-x} ).