Math 22 Derivatives of Exponential Functions

Derivative of the Natural Exponential Function

 Let $u$ be a differentiable function of $x$. Then,
 1.${\frac {d}{dx}}[e^{x}]=e^{x}$
 2.Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d}{dx}[e^u]=e^u\frac{du}{dx}}

Exercises Differentiate each function:

a) $f(x)=e^{2x}$

Solution:
$f'(x)=2e^{2x}$

b) $f(x)=e^{3x^{2}}$

Solution:
$f'(x)=6xe^{3x^{2}}$

c) $f(x)=e^{-x^{2}}$

Solution:
$f'(x)=-2xe^{2x}$

d) $f(x)=4e^{-x}$

Solution:
$f'(x)=-4e^{-x}$

e) $f(x)={\frac {e^{x}-e^{-x}}{2}}$

Solution:
$f(x)={\frac {e^{x}-e^{-x}}{2}}={\frac {e^{x}}{2}}-{\frac {e^{-x}}{2}}={\frac {1}{2}}e^{x}-{\frac {1}{2}}e^{-x}$
$f'(x)={\frac {1}{2}}e^{x}-{\frac {1}{2}}(-1)e^{-x}={\frac {1}{2}}e^{x}+{\frac {1}{2}}e^{-x}$

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