# How do I find the derivative of 3e^(-12t) ?

Jan 2, 2016

You can use the chain rule.

$\left(3 {e}^{- 12 t}\right) ' = - 36 \cdot {e}^{- 12 t}$

#### Explanation:

The 3 is a constant, it can be kept out:

$\left(3 {e}^{- 12 t}\right) ' = 3 \left({e}^{- 12 t}\right) '$

It's a mixed function. The outer function is the exponential, and the inner is a polynomial (sort of):

$3 \left({e}^{- 12 t}\right) ' = 3 \cdot {e}^{- 12 t} \cdot \left(- 12 t\right) ' =$

$= 3 \cdot {e}^{- 12 t} \cdot \left(- 12\right) = - 36 \cdot {e}^{- 12 t}$

Deriving:

If the exponent was a simple variable and not a function, we would simply differentiate ${e}^{x}$. However, the exponent is a function and should be transformed. Let $\left(3 {e}^{- 12 t}\right) = y$ and $- 12 t = z$, then the derivative is:

$\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{\mathrm{dy}}{\mathrm{dt}} \cdot \frac{\mathrm{dz}}{\mathrm{dz}} = \frac{\mathrm{dy}}{\mathrm{dz}} \cdot \frac{\mathrm{dz}}{\mathrm{dt}}$

Which means you differentiate ${e}^{- 12 t}$ as if it were ${e}^{x}$ (unchanged), then you differentiate $z$ which is $- 12 t$ and finally you multiply them.