Matheno - Learn Well and Excel

Derivative of ln(x)

The derivative of the natural log, $\ln(x)$ is
Derivative of ln(x)

\[\bbox[yellow,5px]{\dfrac{d}{dx}\ln x = \dfrac{1}{x} }\]
Applying the Chain rule, we have
\[\dfrac{d}{dx}\ln(\text{stuff}) = \dfrac{1}{(\text{stuff})} \cdot \dfrac{d}{dx} (\text{stuff}) \]
You’ll usually see this written as
\[\dfrac{d}{dx}\ln u = \dfrac{1}{u} \cdot \dfrac{du}{dx} \]

Practice problems are of course below!

Problem #1
Differentiate $f(x) = \ln (x^2 + 3x -1).$
Show/Hide Solution
\begin{align*} f'(x) &= \frac{d}{dx} \left[ \ln (x^2 + 3x -1) \right] \\[8px] &= \frac{1}{(x^2 + 3x -1)} \cdot \left[ \frac{d}{dx}(x^2 + 3x -1) \right] \\[8px] &= \frac{2x + 3}{(x^2 + 3x -1)} \quad \cmark \end{align*}
[hide solution]
Problem #2
Differentiate $f(x) = \sqrt{\ln x^2}.$
Show/Hide Solution
\begin{align*} f'(x) &= \frac{d}{dx} \left[ \sqrt{\ln x^2}\right]\\[8px] &= \dfrac{d}{dx} \left[ \ln x^2\right]^{1/2} \\[8px] &=\frac{1}{2} \left[ \ln x^2\right]^{-1/2} \cdot \left[ \frac{d}{dx}\ln x^2\right] \\[8px] &=\frac{1}{2} \frac{1}{\sqrt{\ln x^2}} \cdot \left[ \frac{1}{x^2} \left(\frac{d}{dx} x^2 \right)\right] \\[8px] &= \frac{1}{2} \frac{1}{\sqrt{\ln x^2}} \cdot \frac{1}{x^2} (2x) \\[8px] &= \frac{1}{x \sqrt{\ln x^2}} \quad \cmark \end{align*}
[hide solution]
Problem #3
Differentiate $f(x) = \ln(\ln x).$
Show/Hide Solution
\begin{align*} f'(x) &= \frac{d}{dx} \left[ \ln(\ln x)\right] \\[8px] &= \frac{1}{\ln x} \cdot \frac{d}{dx}\left( \ln x \right) \\[8px] &= \frac{1}{\ln x} \cdot \frac{1}{x} \\[8px] &= \frac{1}{x \ln x} \quad \cmark \end{align*}
[hide solution]
Problem #4
Differentiate $f(x) = \ln(x^2 e^x).$
Show/Hide Solution
We’ll solve this two ways:
Method 1: Chain rule and Product rule
\begin{align*} \dfrac{d}{dx}\ln(x^2 e^x) &= \dfrac{1}{x^2 e^x} \cdot \dfrac{d}{dx} \left[ x^2 e^x\right] \\[8px] &= \dfrac{1}{x^2 e^x} \cdot \left[\left(\dfrac{d}{dx}x^2 \right)e^x + x^2 \left( \dfrac{d}{dx} e^x\right) \right] \\[8px] &= \dfrac{1}{x^2 e^x} \cdot \left[2x e^x + x^2 e^x \right] \\[8px] &= \dfrac{2x e^x}{x^2 e^x} + \dfrac{x^2 e^x}{x^2 e^x} \\[8px] &= \dfrac{2}{x} + 1 \quad \cmark \end{align*} Method 2: Use log property.
Recall that \begin{align*} \ln(f*g) &= \ln(f) + \ln(g) \\[8px] \text{Hence} \phantom{\ln(f*g)}& \\[8px] \ln(x^2 e^x) &= \ln(x^2) + \ln(e^x) \end{align*} Then we can compute the derivative: \begin{align*} \dfrac{d}{dx}\ln(x^2 e^x) &= \dfrac{d}{dx}\ln(x^2) + \dfrac{d}{dx}\ln(e^x) \\[8px] &= \dfrac{1}{x^2} \cdot \left(\dfrac{d}{dx}x^2 \right) + \dfrac{1}{e^x} \cdot \left(\dfrac{d}{dx}e^x \right) \\[8px] &= \dfrac{1}{x^2} \cdot 2x + \dfrac{1}{e^x} \cdot e^x \\[8px] &= \dfrac{2}{x} + 1 \quad \cmark \end{align*}
[hide solution]
Problem #5
Differentiate $f(x) = \ln\left(\dfrac{\cos x}{5x} \right).$
Show/Hide Solution
We’ll solve this two ways.
Method 1: Chain rule and Quotient rule.
\begin{align*} \dfrac{d}{dx}\left[\ln\left(\dfrac{\cos x}{5x} \right) \right] &= \dfrac{1}{\frac{\cos x}{5x}} \cdot \dfrac{d}{dx}\left[\dfrac{\cos x}{5x} \right] \\[8px] &= \dfrac{5x}{\cos x} \cdot \left[\dfrac{\left(\dfrac{d}{dx}\cos x \right)5x – \cos x \left(\dfrac{d}{dx}5x \right) }{(5x)^2} \right] \\[8px] &= \dfrac{5x}{\cos x} \cdot \left[\dfrac{(-\sin x)5x – \cos x (5)}{(5x)^2} \right] \\[8px] &= \dfrac{1}{\cos x} \left[\dfrac{-5x \sin x – 5 \cos x}{5x} \right] \\[8px] &= \dfrac{-5x\sin x}{5x\cos x} – \dfrac{5 \cos x}{5x \cos x} \\[8px] &= -\tan x – \dfrac{1}{x} \quad \cmark \end{align*} Method 2: Use log property
Recall that Recall that \begin{align*} \ln\left(\dfrac{f}{g} \right) &= \ln(f) – \ln(g) \\[8px] \text{Hence} \phantom{\ln(f*g)}& \\[8px] \ln\left(\dfrac{\cos x}{5x} \right) &= \ln(\cos x) – \ln(5x) \end{align*} Then we can compute the derivative: \begin{align*} \dfrac{d}{dx}\left[\ln\left(\dfrac{\cos x}{5x} \right) \right] &= \dfrac{d}{dx}\ln(\cos x) – \dfrac{d}{dx}\ln(5x) \\[8px] &= \dfrac{1}{\cos x} \cdot \left(\dfrac{d}{dx}\cos x \right) – \dfrac{1}{5x} \cdot \left(\dfrac{d}{dx}5x \right) \\[8px] &= \dfrac{1}{\cos x} \cdot (-\sin x) – \dfrac{1}{5x} \cdot (5) \\[8px] &= -\tan x – \dfrac{1}{x} \quad \cmark \end{align*}
[hide solution]

Please let us know on our Forum:
  • What questions do you have about the solutions above?
  • Which ones are giving you the most trouble?
  • What other problems are you trying to work through for your class about calculating derivatives that you don't know how to do easily yet?
If you let us know, we'll do our best to help!
As of September 2022, we’re using our Forum for comments and discussion of this topic, and for any math questions. We’d love to see you there! Please tap to visit our Forum:

You can support our work with a cup of coffee

We're a small, self-funded team on a mission to provide high-quality, helpful materials to anyone working to learn Calculus well. We provide this site ad-free(!), and we do not sell your data to anyone. We’ve spent thousands of hours crafting everything that's here, and with your help, we can keep growing and offer more!

If we've saved you some time, or you've found our materials useful, please consider giving whatever amount feels right to you. This can take less than 60 seconds, and anything you give helps . . . and if you can contribute a little more, that allows us to continue to provide for those with less.

We thank you in advance for whatever you choose to give.

Yes, I'll give back via PayPal!
Other payment options
(including Google Pay and Apple Pay)

Payment information is completely secure and never touches our servers.

Thank you! ❤️