# How do you find the fourth derivative of e^(-x)?

Apr 19, 2016

${d}^{4} / \left({\mathrm{dx}}^{4}\right) f \left(x\right) = {\left(- 1\right)}^{4} \cdot {e}^{- x} = {e}^{- x}$

#### Explanation:

The best way to take a high number of derivatives of a function is to look for a "generating" function. The derivative of an exponential function is the perfect candidate for this since the derivative is just the function itself times the derivative of the exponent (using the chain rule). For example, the first derivative is:

$\frac{d}{\mathrm{dx}} f \left(x\right) = {e}^{- x} \cdot \frac{d}{\mathrm{dx}} \left(- x\right) = {e}^{- x} \cdot \left(- 1\right)$

Taking the derivative again, we get:

${d}^{2} / \left({\mathrm{dx}}^{2}\right) f \left(x\right) = {e}^{- x} \cdot \left(- 1\right) \cdot \frac{d}{\mathrm{dx}} \left(- x\right) = {e}^{- x} \cdot \left(- 1\right) \cdot \left(- 1\right)$

Notice that each time we take the derivative we get a factor of $- 1$. We can generalize this to:

${d}^{n} / \left({\mathrm{dx}}^{n}\right) f \left(x\right) = {\left(- 1\right)}^{n} \cdot {e}^{- x}$

So the 4th derivative ($n = 4$) is

${d}^{4} / \left({\mathrm{dx}}^{4}\right) f \left(x\right) = {\left(- 1\right)}^{4} \cdot {e}^{- x} = {e}^{- x}$