Find the derivatives of the functions y=x3ex

Question

ottcomn · Accepted Answer

Step 1
We find the first few derivatives using the product rule

y = x^{3} e^{x}

y^{(1)} = 3 x^{2} e^{x} + x^{3} e^{x}

y^{(2)} = 6 x e^{x} + 3 x^{2} e^{x} + 3 x^{2} e^{x} + x^{3} e^{x} = 6 x e^{x} + 6 x^{2} e^{x} + x^{3} e^{x}

y^{(3)} = 6 e^{x} + 6 x e^{x} + 12 x e^{x} + 6 x^{2} e^{x} + 3 x^{2} e^{x} + x^{3} e^{x} = 6 e^{x} + 18 x e^{x} + 9 x^{2} e^{x} + x^{3} e^{x}

y^{(4)} = 6 e^{x} + 18 e^{x} + 18 x e^{x} + 18 x e^{x} + 9 x^{2} e^{x} + 3 x^{2} e^{x} + x^{3} e^{x} = 24 e^{x} + 36 x e^{x} + 12 x^{2} e^{x} + x^{3} e^{x}

Step 2

x^{3} e^{x}

is fixed for all

x^{2} e^{x}

has coefficients = 3,6,9,12. So it follows the pattern 3n

x e^{x}

has coefficients = 6,18, 36. So it follows the pattern 3n(n-1)

e^{x}

has coefficients = 6, 24 etc . It follows the pattern n(n-1)(n-2)
Then we find the nth derivative using the same pattern.

y^{(n)} = x^{3} e^{x} + 3 n x^{2} e^{x} + 3 n (n - 1) x e^{x} + n (n - 1) (n - 2) e^{x}

For n>2
Answer:

y^{(n)} = x^{3} e^{x} + 3 n x^{2} e^{x} + 3 n (n - 1) x e^{x} + n (n - 1) (n - 2) e^{x}

Find the derivatives of the functions y =x^{3}e^{x}