Skip to main content

Section 3.6 Derivatives of Logarithmic Functions

We have already seen that the natural logarithm has derivative

\begin{equation*} \frac{d}{dx}\ln x = \frac{1}{x}, \; x>0 \end{equation*}
Further, for \(x<0\text{,}\) \((\ln |x|)' = (\ln(-x))' = \displaystyle \frac{1}{-x} \cdot (-1) = \frac{1}{x}\text{,}\) so
\begin{equation} \frac{d}{dx}\ln |x| = \frac{1}{x}, \; x \neq0\label{dlogaxdx}\tag{3.6.1} \end{equation}

Differentiate \(y=\ln(x^3+1)\text{.}\)

Find \(\displaystyle\frac{d}{dx}\ln(\sin x)\text{.}\)

Always remember to look simplify first, before differentiating:

\(f(x)=\sqrt{\ln x}\)

Here even more, simplify first!

Find \(\displaystyle\frac{d}{dx}\ln\frac{x+1}{\sqrt{x-2}}\text{.}\)

Other Logarithmic Functions (Rarely Needed!).

Other logarithmic functions are easily converted into terms of the natural logarithm using

\begin{equation*} \log_{a}x = \frac{\ln x}{\ln a} = \frac{1}{\ln a} \ln x, \quad\text{so} \end{equation*}
all logarithmic functions are constant multiples of the natural logarithm. The constant multiple rule then gives their derivatives:
\begin{equation*} \frac{d}{dx}\log_{a} x = \frac{1}{x\ln a}, \; x>0 \end{equation*}
\begin{equation*} \frac{d}{dx}\log_{a} |x| = \frac{1}{x\ln a}, \quad x \neq 0. \end{equation*}
\(f(x)=\log_{10}(2+\sin x)\)
\(a^{(\log_a x)}=x\)(3.6.1)
\begin{equation*} \frac{d}{dx}\log_{a} x = \frac{1}{x\ln a}, \; x>0. \end{equation*}
Two Useful Derivative Formulas.

The Chain Rule gives

\begin{equation} \frac{d}{dx}\ln f(x) = \frac{{df}/{dx}}{f(x)}, \quad \text{ or } \quad (\ln u)' = \frac{u'}{u}.\label{logchainrule}\tag{3.6.2} \end{equation}

One common special case is when function \(u\) is linear:

\begin{equation} \frac{d}{dx}\ln(mx+a) = \frac{m}{mx+a},\label{dlnmxadx}\tag{3.6.3} \end{equation}
and even more specifically,
\begin{equation} \frac{d}{dx}\ln(x+a) = \frac{1}{x+a}.\label{dlnxadx}\tag{3.6.4} \end{equation}

Warning: This is the only case where the derivative of \(\ln f(x)\) is \(1/f(x)\text{!}\)

Having \(f'(x)=1\) is the key.

Verify that the derivative of \(\ln(|\sec x|)\) is \(\tan x\text{.}\)

Solution
Note that \(\displaystyle\sec x = \frac{1}{\cos x}\) and \(\ln(1/u) = -\ln u\text{,}\) so
\begin{equation*} \ln(|\sec x|) = \ln\left(\frac{1}{|\cos x|}\right) = -\ln(|\cos x|). \end{equation*}
Then using the Chain Rule with \(u=\cos x\text{,}\)
\begin{equation*} \frac{d}{dx}\ln(|\sec x|) = -\frac{d}{du}(\ln |u|) \cdot \frac{du}{dx} = -\frac{1}{u} \cdot (-\sin x) = \frac{\sin x}{\cos x} = \tan x. \end{equation*}
Logarithmic Differentiation.

Logarithms have the nice property of converting products to sums, quotients to differences and exponentials to products. The leads to the method of logarithmic differentiation, which can simplify the differentiation of functions built of products, quotients and exponentials.

Compute \(\displaystyle\frac{dy}{dx}\) for \(\displaystyle y=\frac{x^3}{(2x+3)^5}\text{.}\)

Solution

For \(\displaystyle y=\frac{x^3}{(2x+3)^5}\text{,}\) \(\displaystyle\ln y = \ln\frac{x^3}{(2x+3)^{5}} = \ln(x^3)-\ln((2x+3)^5) = 3 \ln x-5\ln(2x+3)\text{.}\)

The left side has derivative \(\displaystyle \frac{d}{dx}{\ln y} = \frac{d}{dy}(\ln y) \frac{dy}{dx} = \frac{1}{y} \frac{dy}{dx}.\)

Using Eq. (3.6.3), the right side has derivative

\begin{equation*} \frac{d}{dx}[3 \ln(x)-5\ln(2x+3)] = 3\frac{1}{x}-5\frac{2}{2x+3} = \frac{3}{x}-\frac{10}{2x+3}. \end{equation*}

Comparing the two sides, \(\displaystyle\frac{1}{y}\frac{dy}{dx} = \frac{3}{x} - \frac{10}{2x+3}.\)

Finally, multiplying each side by \(y\text{,}\)

\begin{equation*} \frac{dy}{dx} = \left(\frac{3}{x} - \frac{10}{2x+3}\right)y = \left(\frac{3}{x} - \frac{10}{2x+3}\right)\frac{x^3}{(2x+3)^5}. \end{equation*}
Procedure for Logarithmic Differentiation.
Starting with a function given as \(y=f(x)\) where the formula for \(f\) is a term built with products, quotients and/or powers.
  1. Write \(\ln y = \ln (\cdots)\text{,}\) and simplify the right hand side as much as possible, using the laws for the logarithms of products, quotients and/or powers.
  2. Compute the derivative of each side, probably using the above special case of the Chain Rule \((\ln u)' = \displaystyle\frac{u'}{u}.\) This gives an equation
    \begin{equation*} (\ln y)' = \frac{y'}{y} = \cdots \end{equation*}
  3. Multiply each side by \(y\text{,}\) inserting the formula \(f(x)\) for \(y\) on the right, giving
    \begin{equation*} y' = f(x)[\cdots]. \end{equation*}
\(\displaystyle y=\frac{x^{3/4}\sqrt{x^2+1}}{(3x+2)^5}\)
Sometimes, logarithmic differentiation is not just a shortcut, but the only obvious method:
\(y=x^{\sqrt{x}}\)
The Number \(e\) as a Limit.
We will not cover this in class, or on tests, but read it as an example. One can get the formula
\begin{equation*} e = \lim_{n \to \infty} \left(1+\frac{1}{n}\right)^n \end{equation*}
which can be used to approximate \(e\) by evaluating for a large value of \(n\text{.}\)

Recommended Exercises Recommended Exercises

Study Exercises 2, 3, 11, 12, 19, 23, 27, 39, 43, 44 and 45 from the text.