05 7expfuns

The derivative of ex

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Given $f(x)=e^x$

… … $f'(x)=e^x$

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For $f(x)=e^{kx}, \quad k \in R$

… … $f'(x)=ke^{kx}$

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05.7eg1.gif

Example 1

Differentiate $y=e^{-2x}$

Solution:

… … $\dfrac{dy}{dx}=-2e^{-2x}$

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Example 2

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

Solution:

05.7eg2.gif

… … $y=\dfrac{1}{e^{2x}+1}$

… … Use the chain rule.

… … Let $u=e^{2x}+1 \quad \text{ then } y=\dfrac{1}{u}$

… … $\dfrac{du}{dx}=2e^{2x} \qquad \quad \dfrac{dy}{du}=-u^{-2}$

… … Now use

… … $\dfrac{dy}{dx}=\dfrac{dy}{du}.\dfrac{du}{dx}$

… … $\dfrac{dy}{dx}=\dfrac{-1}{u^2}\times2e^{2x}$

… … … . $=\dfrac{-2e^{2x}}{(e^{2x}+1)^2}$

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Example 3

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

Solution:

05.7eg3.gif

… … Simplify

… … $y = \dfrac{e^{2x}+e^{-4x}} {e^x}$

… … $y=\dfrac{e^{2x}}{e^x}+ \dfrac{e^{-4x}}{e^x}$

… … $y=e^x+e^{-5x}$

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… … hence

… … $\dfrac{dy}{dx}=e^x-5e^{-5x}$

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

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Example 4

Find the derivative of $y=e^{(1-3x)^3}$

Solution:

05.7eg4.gif

… … $y=e^{(1-3x)^3}$

… … Use the chain rule:

… … Let $u=(1-3x)^3 \qquad \text{ then } \quad y=e^u$

… … $\dfrac{du}{dx}=3 \times (1-3x)^2 \times \big( -3 \big) \qquad \dfrac{dy}{du}=e^u$

… … $\dfrac{dy}{dx}=e^u\times-9(1-3x)^2$

… … … $=-9(1-3x)^2 \, e^{(1-3x)^3}$

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

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