Derivatives of Logarithms of Functions

\\ \t $\dfrac{d}{dx}(e^x) = e^x$ \gap[20] \t Yes, the derivative of $e^x$ is itself! \\ \t $\dfrac{d}{dx}(b^x) = b^x \ln b \quad$ \t If $b = e$ then $\ln b = 1$, so this agrees with the one above.

#[Applying the chain rule gives][Aplicar la regla de la cadena da]#

\\ \t $\dfrac{d}{dx}[e^{\color{blue}{u}}] = e^{\color{blue}{u}}\dfrac{d\color{blue}{u}}{dx}$ \gap[20] \t #[The derivative of $e$ raised to a function][La derivada de $e$ elevada a una función]# \\ \t $\dfrac{d}{dx}[b^{\color{blue}{u}}] = b^{\color{blue}{u}} \ln b\dfrac{d\color{blue}{u}}{dx} \quad$ \t #[The derivative of $b$ raised to a function][La derivada de $b$ elevada a una función]#

#[Examples][Ejemplos]#

\\ $\dfrac{d}{dx}\left[e^{x^2+x}\right] = e^{x^2+x}\dfrac{d}{dx}[x^2+x]$ \t ${}= (2x+1)e^{x^2+x}$

\\ $\dfrac{d}{dx}\left[b^{x^2+x}\right] = b^{x^2+x}\ln b\dfrac{d}{dx}[x^2+x]$ \t ${}= (2x+1)b^{x^2+x}\ln b$

#[Some for you][Algunos para ti]#

Calculate the given derivative.

$\displaystyle \frac{d}{d%0}\left[%2\right]$

= BOX*

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$\displaystyle \frac{d}{d%3}\left[%5\right]$

= BOX*

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