Math 22 Derivatives of Exponential Functions

Derivative of the Natural Exponential Function

 Let Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u}
 be a differentiable function of $x$. Then,
 1.Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {d}{dx}}[e^{x}]=e^{x}}

 2.${\frac {d}{dx}}[e^{u}]=e^{u}{\frac {du}{dx}}$

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

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

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

Solution:
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 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}}
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 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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