## Tutorial: Derivatives of logarithmic and exponential functions

This tutorial: Part B: Derivatives of exponential functions

(This topic is also in Section 4.5 in *Applied Calculus*or Section 11.5 in

*Finite Mathematics and Applied Calculus*) #[I don't like this new tutorial. Take me back to the old tutorial!][No me gusta este nueve tutorial. ¡Regresame al tutorial más viejo!]#

#[Derivative of][Derivada de]#

The derivative of $e^x$ and $b^x$ are given by the following formulas
*e*#[and][y]#^{x}*b*^{x}**1. Derivative of $e^x$**

$\dfrac{d}{dx}(e^x) = e^x$ \gap[20] \t Yes, the derivative of $e^x$ is itself!

**2. Derivative of $b^x$**

$\dfrac{d}{dx}(b^x) = b^x \ln b$ \gap[1] \t When $b = e$, this agrees with (1); see below.

**#[Note][Nota]#**When $b = e$ in formula (2), then $\ln b = \ln e = 1$, so (2) gives the same formula as (1).

**%%Examples**

**1.**$\dfrac{d}{dx}(5^x) = 5^x\ln\,5$ \gap[5] \t

#[Formula][Fórmula]# 2

\\ **2.**$\dfrac{d}{dx}(5e^x) = 5e^x$ \gap[5] \t Derivative of a constant times a function \\

**3.**$\dfrac{d}{dx}\left[-4(20^x)\right] = -4(20^x)\ln\,20$ \gap[5] \t Derivative of a constant times a function \\

**4.**$\dfrac{d}{dx}\left(x^2e^x) = 2xe^x + x^2e^x$ \gap[5] \t Product rule \\ $\qquad {}= e^x(2x+x^2)$

**Some for you**

**%%Q**Where do these formulas come from?

**%%A**For derivations of these formulas, consult Section 4.5 in

*Applied Calculus*or Section 11.5 in

*Finite Mathematics and Applied Calculus*.

Derivatives of exponentials of functions

We now know how to differentiate expressions that contain $b^x$ for any base $b$. What about $b$ raised to more complicated quantities, for instance $2^{x^2-3x+2}$? For things like this we need to use the chain rule (see the %%chainruletut):
**Differentiating exponentials of functions**

\\ \t
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\t
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*Derivative of $e$ rised to a quantity is $e$ raised to that that quantity times the derivative of that quantity.*

*The derivative of the b raised to a quantity is the product of b raised to that quantity with ln b, times the derivative of that quantity.*

**%%Examples**

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

\\

**6.**$\dfrac{d}{dx}\left[3^{\color{blue}{x^2+x}}\right] = \left(3^{\color{blue}{x^2+x}}\ln 3\right)\dfrac{d}{dx}[\color{blue}{x^2+x}]$ \t ${}= (2x+1)3^{x^2+2}\ln 3$**Some for you**

Now try the exercises in Section 4.5 in

*Applied Calculus*or Section 11.5 in*Finite Mathematics and Applied Calculus*.*January 2022*

Copyright © 2019 Stefan Waner and Steven R. Costenoble

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