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# Derivative of ln – Derivation, Explanation, and Example

Weâ€™ll focus on understanding the **derivative of ln** in our discussion. The natural logarithmic function helps us quantify and model values that are significantly small or large. With its extensive application in physics and biological sciences, itâ€™s important to know how to differentiate natural logarithmic functions.

*The derivative of ln returns the reciprocal of the input value. Whenever possible, the chain rule still applies.*

In this article, weâ€™ll learn how to apply the derivative rule for functions that contain $\ln$ in their expression. Weâ€™ll also understand how we can use $y = e^x$, its inverse function, to derive the formula for the derivative of $\ln x$.

**What is the derivative of ln?Â Â Â Â **

The derivative of $\ln$ is one of the most remembered derivative rules. Thatâ€™s because the **derivative of** $\boldsymbol{\ln x}$ **is simply the reciprocal of** $\boldsymbol{x}$. Here are two graphs showing us how $y = \ln x$ and $yâ€™ = \dfrac{1}{x}$ would look like on an $xy$-plane.

We can extend this rule for composite functions that contain $\ln$ as its outer function.

\begin{aligned}\dfrac{d}{dx}\ln x &= \dfrac{1}{x}\\\dfrac{d}{dx} \ln [g(x)] &= \dfrac{1}{g(x)} \cdot \dfrac{d}{dx} g(x)\\&= \dfrac{gâ€™(x)}{g(x)}\end{aligned}

This means that the chain rule will still apply when we differentiate composite functions that implement the derivative rule for $\ln$.

**How to find the derivative of ln and functions containing it?**

The derivative of $\ln$ shows us that itâ€™s possible to end up with a rational expression when differentiating functions that are seemingly complex such as $\ln x$. This derivative rule, $\dfrac{d}{dx} \ln x = \dfrac{1}{x}$, will come in handy once we learn how to integrate functions. For now, letâ€™s take a look at how we were able to derive this rule.

**What is the derivative of ln xâ€™s proof?**

We can begin by applying the rules of logarithms. Recall that we can write $e^{\ln x}$ as $x$. Letâ€™s see what happens when we differentiate both sides of the equations.

\begin{aligned}e^y &= x\\\dfrac{dy}{dx} e^y &= \dfrac{dy}{dx} x\\\dfrac{dy}{dx}e^y &= 1\end{aligned}

Letâ€™s substitute $e^y$ with $x$ to rewrite the equation’s left-hand side in terms of $x$. We can then divide both sides by $x$.

\begin{aligned}e^{\ln x} &= x\\\dfrac{d}{dx} e^{\ln x} &= \dfrac{d}{dx} x\\\dfrac{d}{dx} e^{\ln x} &={\color{Purple} 1}, \color{Purple}\text{Power Rule}\end{aligned}

Recall that $\dfrac{d}{dx} e^x = e^x$, so we can use this to simplify the left-hand side expression. Make sure to apply the chain rule when differentiating $e^{\ln x}$.

\begin{aligned}\dfrac{d}{dx} e^{\ln x} &= 1\\ {\color{Purple}(e^{\ln x})} \cdot {\color{DarkBlue}\dfrac{d}{dx} \ln x} &= 1,\phantom{x}{\color{Purple}\text{Derivative of }e^x}\text{ & }{\color{DarkBlue}\text{Chain Rule}}\end{aligned}

Replace $e^{\ln x}$ with $x$ then divide both sides of the equation by $x$ to isolate $\dfrac{d}{dx} \ln x$ on the left-hand side of the equation.

\begin{aligned}e^{\ln x} \dfrac{d}{dx} \ln x &= 1\\x \dfrac{d}{dx}\ln x &= 1\\\dfrac{d}{dx}\ln x&= \dfrac{1}{x}\end{aligned}

Thus, proving that $\dfrac{d}{dx} \ln x =\dfrac{1}{x}$ is indeed true. Â Now that weâ€™ve shown you how to prove this rule letâ€™s see how we can apply this derivative rule to differentiate other functions.

**How to use the derivative rule of ln x?**

When given a function or a composite function that contains $\ln x$ or $\ln[g(x)]$, we can apply the following process to differentiate the functions:

- Take note of the variable or the inner function.
- Take the reciprocal of the variable or inner function.
- When working with a composite function, make sure to differentiate the inner function.

Letâ€™s say we want to differentiate $y = \ln (x +3)$; we take the reciprocal of $(x + 3)$ and account for the inner functionâ€™s derivative through the chain rule.

\begin{aligned}\dfrac{d}{dx} \ln (x +3) &= {\color{Purple}\dfrac{1}{x +3}} \cdot {\color{DarkBlue} \dfrac{d}{dx}(x +3)}, \phantom{x}{\color{Purple}\text{Derivative of ln}}\text{ & }{\color{DarkBlue} \text{Chain Rule}}\\&= \dfrac{1}{x +3}\cdot \left({\color{Purple}\dfrac{d}{dx} x + \dfrac{d}{dx} 3} \right ),\phantom{x}{\color{Purple}\text{Sum Rule}}\\&= \dfrac{1}{x +3}\cdot ({\color{Purple}1}+{\color{DarkBlue}0})\phantom{x}{\color{Purple}\text{Power Rule}}\text{ & }{\color{DarkBlue} \text{Constant Rule}}\\&= \dfrac{1}{x + 3}\end{aligned}

This shows how easy it is for us to differentiate functions with $\ln x$ in their expression or outer function. Of course, the fundamental derivative rules will still apply, so make sure to have a list handy when working on the examples shown below.

*Example 1*

Find the derivative of the function, $f(x)= 4\ln(2x^2 – 1)$.

__Solution__

To find $fâ€™(x)$, we can begin by taking $4$ out from the expression that needs to be differentiated. We can then apply the derivative rule, $\dfrac{d}{dx} \ln x = \dfrac{1}{x}$, and the chain rule to differentiate the remaining expression.

\begin{aligned}f'(x) &= \dfrac{d}{dx} 4\ln (2x^2 – 1)\\&= 4\dfrac{d}{dx}\ln (2x^2 -1),\phantom{x}{\color{Purple} \text{Constant Multiple Rule}}\\&= 4 \left({\color{Purple}\dfrac{1}{2x^2 -1}} \right ) \cdot {\color{DarkBlue}\dfrac{d}{dx} (2x^2 – 1)},\phantom{x}{\color{Purple} \text{Derivative of }\ln x}\text{ & }{\color{DarkBlue}\text{Chain Rule}}\end{aligned}

Differentiate $(2x^2 -1)$ using our fundamental derivative rules to further simplify $fâ€™(x)$.

\begin{aligned}f'(x) &= 4 \left(\dfrac{1}{2x^2 -1} \right ) \cdot \left( {\color{Purple}\dfrac{d}{dx} 2x^2 – \dfrac{d}{dx}1}\right),\phantom{x}{\color{Purple} \text{Difference Rule}}\\&= \dfrac{4}{2x^2 -1}\cdot \left( {\color{Purple}2\dfrac{d}{dx} x^2 }- {\color{DarkBlue}0}\right),\phantom{x}{\color{Purple} \text{Constant Multiple Rule}}\text{ & }{\color{DarkBlue}\text{Chain Rule}}\\&= \dfrac{4}{2x^2 -1}\cdot \left( {\color{Purple}2(2x)}\right),\phantom{x}{\color{Purple} \text{Power Rule}}\\&= \dfrac{16x}{2x^2 -1 }\end{aligned}

Hence, we have $fâ€™(x)= \dfrac{16x}{2x^2 -1 }$.

*Example 2*

Find the derivative of the function, $g(x)= \ln [(x -1)(x + 4)]^4$.

__Solution__

Before we evaluate $gâ€™(x)$, letâ€™s rewrite the expression using the logarithmic property, $\ln a^n = n \ln a$.

\begin{aligned}g(x) &= 4\ln [(x -1)(x + 4)]\end{aligned}

Differentiate $g(x)$ by using the derivative rule for $\ln x$ and apply the chain rule to account for the derivative of $(x -1)(x +4)$.

\begin{aligned}g'(x) &= \dfrac{d}{dx} 4\ln [(x -1)(x + 4)]\\&= 4\dfrac{d}{dx}\ln [(x -1)(x + 4)],\phantom{x}{\color{Purple}\text{Constant Multiple Rule}}\\&=4\cdot{\color{Purple}\dfrac{1}{(x -1)(x +4)}}\cdot {\color{DarkBlue} \dfrac{d}{dx}(x -1)(x + 4)},\phantom{x}{\color{Purple}\text{Derivative of }\ln x}\text{ & }{\color{DarkBlue}\text{Chain Rule}}\end{aligned}

Â Apply the product rule to evaluate $\dfrac{d}{dx} (x -1)(x + 4)$.Â Letâ€™s focus on this expression first and go back to $gâ€™(x)$ once we simplify the derivativeâ€™s expression.

\begin{aligned} \dfrac{d}{dx}(x -1)(x + 4) &= (x -1)\dfrac{d}{dx} (x +4) + (x+ 4)\dfrac{d}{dx}(x- 1),\phantom{x} {\color{Purple}\text{Product Rule}}\\&= (x -1)\left({\color{Purple}\dfrac{d}{dx} x+\dfrac{d}{dx}4} \right ) + (x+4)\left({\color{Purple}\dfrac{d}{dx} x-\dfrac{d}{dx}1} \right )\phantom{x} {\color{Purple}\text{Sum & Difference Rules}}\\&= (x -1)({\color{Purple}1} + {\color{DarkBlue}0}) +(x +4)({\color{Purple}1} – {\color{DarkBlue}0}),\phantom{x} {\color{Purple}\text{Power Rule}}\text{ & }{\color{DarkBlue}\text{Constant Rule}}\\&= (x -1) + (x + 4)\\&= 2x +3\end{aligned}

Letâ€™s go back to our current expression for $gâ€™(x)$ and substitute the third factor with $2x + 3$.

\begin{aligned}g'(x) &=4\cdot\dfrac{1}{(x -1)(x +4)}\cdot \dfrac{d}{dx}(x -1)(x + 4)\\&=4\cdot\dfrac{1}{(x -1)(x +4)}\cdot (2x +3)\\&= \dfrac{4(2x + 3)}{(x -1)(x +4)}\end{aligned}

This means that $gâ€™(x) = \dfrac{4(2x + 3)}{(x -1)(x +4)}$.

*Example 3*

The height of a tree after $x$ years since it was planted can be modeled using the function, $h(x) = 8\cdot \ln x$. Â What is the rate at which the tree is growing four years after it was planted?

__Solution__

We can find the growth rate of the tree after four years by evaluating $hâ€™(4)$. We can apply the constant multiple rule and $\dfrac{d}{dx}\ln x=\dfrac{1}{x}$ to find the expression for $hâ€™(x)$.

\begin{aligned}h'(x) &= \dfrac{d}{dx} 8\ln x\\&= 8\dfrac{d}{dx}\ln x,\phantom{x}{\color{Purple}\text{Constant Multiple Rule}}\\&=8\cdot{\color{Purple}\dfrac{1}{x}},\phantom{x}{\color{Purple}\text{Derivative of }\ln x}\\&=\dfrac{8}{x}\end{aligned}

Use this expression to find $hâ€™(4)$. Hence, we have $hâ€™(x) = \dfrac{8}{4} = 2$. This means that the treeâ€™s growth rate is $\boldsymbol{2}$ **feet per year**.

**Practice Questions**

1. Find the derivative of the following functions.

a. $f(x) = 6\ln(2x^2 + 5)$

b. $g(x) = -2\ln(4x^3 -2x)$

c. $h(x) = 4\ln(\sqrt{2x})$

2. Find the derivative of the following functions.

a. $f(x) = \ln[(x -1)(x + 2)]^3$

b. $g(x) = -\ln\left(\dfrac{x + 4}{x – 5}\right)$

c. $h(x) = 3\ln(\sqrt{x^2 + 2x +1})$

3. Find the derivative of the following functions.

a. $f(x) = \ln(\cos^2 x)$

b. $g(x) = 3x\ln^2 x$

c. $h(x) = \dfrac{3\ln x}{\sqrt{x^2 + 4}}$

### Answer Key

1.

a. $fâ€™(x) = \dfrac{24x}{2x^2 + 5}$

b. $gâ€™(x)= -\dfrac{2\left(6x^2 – 1\right)}{x\left(2x^2 – 1\right)}$

c. $hâ€™(x) = \dfrac{2}{x}$

2.

a. $fâ€™(x) = \dfrac{3\left(2x + 1\right)}{\left(x – 1\right)\left(x + 2\right)}$

b. $gâ€™(x)= \dfrac{9}{\left(x+4\right)\left(x-5\right)}$

c. $hâ€™(x) = \dfrac{3}{x + 1}$

3.

a. $fâ€™(x) = -2 \tan x$

b. $gâ€™(x) = 3(\ln^2 x+ 2\ln x)$

c. $hâ€™(x) = \dfrac{3\left[x^2 + 4 – x^2 \ln \left(x\right)\right]}{x\left(x^2 + 4\right)\sqrt{x^2 + 4}}$

*Images/mathematical drawings are created with GeoGebra.*