## Tutorial: Derivatives of logarithmic and exponential functions

* Adaptive game version*

This tutorial: Part A: Derivatives of logarithmic 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]# ln

The derivative of $\ln x$ and $\log_b x$ are given by the following formulas
*x*#[and][y]# log_{b}*x***1. Derivative of the natural logarithm**

$\dfrac{d}{dx}(\ln x) = \dfrac{1}{x}$ \gap[20]

**2. Derivative of the logarithm with base $b$**

$\dfrac{d}{dx}(\log_bx) = \dfrac{1}{x\ln b}$ \gap[5] \t #[When $b = e$, this agrees with (1); see below.][Cuando $b = e$, entonces esto concuerda con (1); vea abajo.]#

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

**%%Examples**

**1.**$\dfrac{d}{dx}(\log_5x) = \dfrac{1}{x\ln\,5}$ \gap[5] \t

#[Formula][Fórmula]# 2

\\ **2.**$\dfrac{d}{dx}(5\,\ln x) = 5\cdot\dfrac{1}{x} = \dfrac{5}{x}$ \gap[5] \t Derivative of a constant times a function \\

**3.**$\dfrac{d}{dx}(-4\,\log_{20} x) = -4\cdot\dfrac{1}{x\ln\,20}$ \gap[5] \t Derivative of a constant times a function \\ $\qquad {}= -\dfrac{4}{x\ln\,20}$ \\

**4.**$\dfrac{d}{dx}\left(x^2\,\ln x\right) = 2x\,\ln x+ x^2\dfrac{1}{x}$ \gap[5] \t Product rule \\ $\qquad {}= 2x\,\ln x+x$

**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 logarithms of functions

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

\\ \t
\\
\t
\\

*Derivative of the natrual logarithm of a quantity is 1 over that quantity times the derivative of that quantity.*

*The derivative of the log to base b of a quantity is 1 over the product of ln b and that quantity, times the derivative of that quantity.*

**%%Examples**

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

\\

**6.**$\dfrac{d}{dx}\log_3 (\color{blue}{x^2+x}) = \dfrac{1}{(\color{blue}{x^2+x})\ln 3}\dfrac{d}{dx}[\color{blue}{x^2+x}]$ \t ${}= \dfrac{2x+1}{(x^2+2)\ln 3}$**Some for you**

Logs of absolute values

Something curious happens if we take the derivative of the logarithm of the absolute value of $x$:
$\dfrac{d}{dx} \ln \big|x\big|$ \t ${}=\dfrac{1}{\big|x\big|} \dfrac{d}{dx}\big|x\big|$ \t #[Chain rule][Regla de la cadena]#
\\ \t ${}=\dfrac{1}{\big|x\big|} \dfrac{\big|x\big|}{x}$ \t #[The derivative of $\big|x\big|$ is $\frac{\big|x\big|}{x}$.][La derivada de $\big|x\big|$ es $\frac{\big|x\big|}{x}$.]#
\\ \t ${}= \dfrac{1}{x}$ \t Exactly the same as the derivative of $\ln x$!

In other words, replacing $x$ by the absolute value of $x$ has absolutely no effect on the derivative of the natural logarithm. Similarly it has no effect on the derivative of the logarithm of $x$ to any base, or, using the chain rule, on the logarithm of any quantity.
**Derivatives of logarithms of absolute values**

When taking the derivative of the logarithm of an absolute value, just pretend the absolute value signs were just parentheses, so that the result has no absolute value signs:

\t
\\
\t
\\

*The derivative of the natrual logarithm of the absolute value of a quantity is 1 over that quantity times the derivative of that quantity.*

*The derivative of the log to base b of the absolute value of a quantity is 1 over the product of ln b and that quantity, times the derivative of that quantity.*

**%%Examples**

**7.**$\dfrac{d}{dx}(\log_5\big|x\big|) = \dfrac{1}{x\ln\,5}$ \gap[5] \t

#[Compare Example][Comparar el ejemplo]# 1 #[above.][arriba.]#

\\ **8.**$\dfrac{d}{dx}(5\,\ln \big|x\big|) = 5\cdot\dfrac{1}{x} = \dfrac{5}{x}$ \gap[5] \t #[Compare Example][Comparar el ejemplo]# 2 #[above.][arriba.]# \\

**9.**$\dfrac{d}{dx}\left(x^2\,\ln \big|x\big|\right) = 2x\,\ln x+ x^2\dfrac{1}{x}$ \gap[5] \t Compare Example 4 above. \\ $\qquad {}= 2x\,\ln x+x$ \\

**10.**$\dfrac{d}{dx}\ln \big|\color{blue}{x^2+x}\big| = \dfrac{1}{\color{blue}{x^2+x}}\dfrac{d}{dx}[\color{blue}{x^2+x}]$ \gap[5] \t Compare Example 5 above. \\ $\qquad {}= \dfrac{2x+1}{x^2+2}$ \\

**11.**$\dfrac{d}{dx}\log_3 \big|\color{blue}{x^2+x}\big| = \dfrac{1}{(\color{blue}{x^2+x})\ln 3}\dfrac{d}{dx}[\color{blue}{x^2+x}]$ \gap[5] \t Compare Example 6 above. \\ $\qquad {}= \dfrac{2x+1}{(x^2+2)\ln 3}$ \\

**12.**$\dfrac{d}{dx}\ln \big|3x+1\big|^5 = \dfrac{d}{dx}5\ln \big|3x+1\big|$ \gap[5] \t Properties of logarithms \\ $\qquad {}= 5\dfrac{1}{3x+1}\cdot 3 = \dfrac{15}{3x+1}$ \t Compare Example 10 above.

**Some for you**

Now try the exercises in Section 4.5 in

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

Copyright © 2019 Stefan Waner and Steven R. Costenoble