How do you find the first and second derivatives of y = (x^2 + 3) / e^-x using the quotient rule?

Nov 2, 2015

$y ' = {e}^{x} \left({x}^{2} + 2 x + 3\right)$
$y ' ' = {e}^{x} \left({x}^{2} + 4 x + 5\right)$

Explanation:

Alternatively, you could rewrite the original question as a product and apply the product rule instead of the quotient rule.

STEP 1: Rewrite the original equation
$y = \frac{{x}^{2} + 3}{e} ^ - x = \left({x}^{2} + 3\right) \cdot {e}^{x}$

STEP 2: Use the product rule to find the first derivative
$y ' = \left(2 x\right) \left({e}^{x}\right) + \left({x}^{2} + 3\right) \cdot {e}^{- x}$

$y ' = 2 x {e}^{x} + {x}^{2} {e}^{x} + 3 {e}^{x} = {e}^{x} \left({x}^{2} + 2 x + 3\right)$
$y ' ' = {e}^{x} \cdot \left({x}^{2} + 2 x + 3\right) + {e}^{x} \cdot \left(2 x + 2\right)$
$= {e}^{x} \left({x}^{2} + 2 x + 3 + 2 x + 2\right)$
$= {e}^{x} \left({x}^{2} + 4 x + 5\right)$