07Eexponentialderivative

toc = The derivative of e x =

math \\ \text{Given } f(x)=e^x \\ \\ . \qquad f'(x)=e^x math

math \\ \text{For } f(x)=e^{kx}, \quad k \in R \\ \\ . \qquad f'(x)=ke^{kx} math

** Example 1 **
math \text{Differentiate}\;y=e^{-2x} math

__Solution:__

math \dfrac{dy}{dx}=-2e^{-2x} math

** Example 2 **
math \text{Differentiate}\;y=\dfrac{1}{e^{2x}+1} math

__Solution:__

math y=\dfrac{1}{e^{2x}+1} math

Use the chain rule. math \\ \text{Let } u=e^{2x}+1 \quad \text{ then } y=\dfrac{1}{u} \\ \\ .\quad\dfrac{du}{dx}=2e^{2x} \qquad \quad \dfrac{dy}{du}=-u^{-2} math

Now use math \dfrac{dy}{dx}=\dfrac{dy}{du}.\dfrac{du}{dx} math

math \\ \dfrac{dy}{dx}=\dfrac{-1}{u^2}\times2e^{2x} \\ \\ . \quad=\dfrac{-2e^{2x}}{(e^{2x}+1)^2} math

** Example 3 **
math \text{Find}\;\dfrac{dy}{dx},\;\text{given that}\;y=\dfrac{e^{2x}+e^{-4x}}{e^x} math

__Solution:__

Simplify math \\ y = \dfrac{e^{2x}+e^{-4x}} {e^x} \\ \\ y=\dfrac{e^{2x}}{e^x}+ \dfrac{e^{-4x}}{e^x} \\ \\ y=e^x+e^{-5x} math

hence math \dfrac{dy}{dx}=e^x-5e^{-5x} math

Notice the different format for the calculator solution can be resolved using the **expand** function (in the Transformation menu)

** Example 4 **
math \text{Find the derivative of } y=e^{(1-3x)^3} math

__Solution:__ math y=e^{(1-3x)^3} math

Use the chain rule: math \\ \text{Let } u=(1-3x)^3 \qquad \text{ then } \quad y=e^u \\ \\ \dfrac{du}{dx}=3 \times (1-3x)^2 \times \big( -3 \big) \qquad \dfrac{dy}{du}=e^u math

math \\ \dfrac{dy}{dx}=e^u\times-9(1-3x)^2 \\ \\ .\quad=-9(1-3x)^2 \, e^{(1-3x)^3} math

Notice the difference in the calculator solution is due to the negative being taken out as a factor from the (1 – 3x) term.

Go to top of page flat .