Calculus: Derivatives I
Exponentials
The derivative of the exponential function is given by the following formula:
$$\dfrac{d}{dx} e^x = e^x$$
Isn't that neat? The derivative of $e^x$ is itself. That little guy's his own rate of change. To see the derivation of this rule, see the Definition section of Derivatives I.
Problems
Differentiate with respect to $x$: $y=e^{x}$
$\dfrac{dy}{dx}y = \dfrac{d}{dx}e^x \\ \dfrac{d}{dx} = e^x$
Just making sure you're paying attention.Differentiate with respect to $x$: $y=2e^{x+1}$
$\dfrac{d}{dx}y = \dfrac{d}{dx} 2e^{x+1} \\ \dfrac{dy}{dx} = \dfrac{d}{dx} 2e^{x}e^{1}\\ \dfrac{dy}{dx} = 2e \dfrac{d}{dx} e^{x}\\ \dfrac{dy}{dx} = 2e \cdot e^{x}\\ \dfrac{dy}{dx} = 2 e^{x+1}\\ $Differentiate with respect to $x$: $y=e^{a \cdot m}$
$\dfrac{d}{dx}y = \dfrac{d}{dx} e^{a \cdot m} \\ \dfrac{dy}{dx} = 0\\ $Differentiate with respect to $x$: $y=\dfrac{1}{e^{-x}}$
$\dfrac{d}{dx}y = \dfrac{d}{dx}\dfrac{1}{e^{-x}} \\ \dfrac{dy}{dx} = \dfrac{d}{dx}e^{x}\\ \dfrac{dy}{dx} = e^{x}\\ $Differentiate with respect to $x$: $y=\sum\limits_{i=1}^{50}\left(e^x+2\right)$
$\dfrac{d}{dx}y = \dfrac{d}{dx}\sum\limits_{i=1}^{50}\left(e^x+2\right) \\ \dfrac{dy}{dx} = \sum\limits_{i=1}^{50}\dfrac{d}{dx}\left(e^x+2\right) \\ \dfrac{dy}{dx} = \sum\limits_{i=1}^{50}\left(\dfrac{d}{dx}e^x+\dfrac{d}{dx}2\right) \\ \dfrac{dy}{dx} = \sum\limits_{i=1}^{50}\left(e^x+ 0 \right) \\ \dfrac{dy}{dx} = \sum\limits_{i=1}^{50}e^x \\ \dfrac{dy}{dx} = 50e^x \\ $