Section 3.9 Derivatives of Exponential and Logarithmic Functions (and Logarithmic Differentiation)
References.
- Calculus, Early Transcendentals by Stewart, Section 3.6.
In this course, we have already seen and worked with the derivatives of exponential functions (in Section 3.6 and of logarithmic functions (in Section 3.7). Thus for us, the first part of this section of the text is just for review and further worked examples and homework exercises, and I will just summarise briefly with some examples. I suggest looking at Theorems 14, 15 and 26 in the text to review the formulas, and its Examples 74, 75, 77, and 79.
The last part of this section introduces a useful new technique, Logarithmic Differentiation, which is a use of implicit differentiation to simplify differentiation of functions that involve a mix of products, quotients and powers.
Derivatives of exponential and logarithmic functions (recap).
The natural exponential function has derivative
\begin{equation}
\frac{d}{dx}e^x = e^x.\tag{3.9.1}
\end{equation}
and more generally
\begin{equation}
\frac{d}{dx}a^x = (\ln a) a^x, \; a > 0\tag{3.9.2}
\end{equation}
The natural logarithm function has derivative
\begin{equation}
\frac{d}{dx}\ln x = \frac{1}{x}, \; x>0\tag{3.9.3}
\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 \neq 0\tag{3.9.4}
\end{equation}
Checkpoint 3.9.2. Find \(\displaystyle\frac{d}{dx}\ln(\sin x)\).
Always remember to look simplify first, before differentiating:
Checkpoint 3.9.3. Differentiate \(f(x)=\sqrt{\ln x}\).
Here even more, simplify first!
Checkpoint 3.9.4. Find \(\displaystyle\frac{d}{dx}\ln\frac{x+1}{\sqrt{x-2}}\).
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}.\tag{3.9.5}
\end{equation}
One common special case is when function \(u\) is linear:
\begin{equation}
\frac{d}{dx}\ln(mx+a) = \frac{m}{mx+a},\tag{3.9.6}
\end{equation}
and even more specifically,
\begin{equation}
\frac{d}{dx}\ln(x+a) = \frac{1}{x+a}.\tag{3.9.7}
\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.
Example 3.9.5. Verify that the derivative of \(\ln(|\sec x|)\) is \(\tan x\).
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.
Example 3.9.6. Compute \(\displaystyle\frac{dy}{dx}\) for \(\displaystyle y=\frac{x^3}{(2x+3)^5}\).
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. (3.9.6), 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*}
The working of this example reveals a strategy for usign logrutmns to simplify teh differentiation of a function \(y=f(x)\) when the formuals for \(\) is butrok fomr products quotients, powers and expontiations.
- First, take the derivative of both sides, getting \(\ln(y) = \ln(f(x))\)
-
Next, a critical step: simplify the right-hand side as much as possible, using the properties of logarithms; in particular:
- \(\ln(a b) = \ln(a) + \ln(b)\text{,}\)
- \(\ln(a/b) = \ln(a) - \ln(b)\text{,}\) and
- \(\ln(a^b) = b \ln(a)\text{.}\)
(Also, as usual, convert roots to fractional powers!) This will give something like\begin{equation*} \ln y = \ln(f_1(x)) + \cdots \end{equation*} - Differentiate both sides of the equation, with much use of the rule \(\displaystyle \frac{d}{d x} (\ln(u)) = \frac{1}{u}\frac{du}{dx}\text{;}\) This gives an equation that starts\begin{equation*} \frac{1}{y}\frac{dy}{dx} = \cdots \end{equation*}
- Multiply both sides by the quantity \(y=f(x)\text{,}\) the original function being differentiated.
Also look at Examples 81 to 83 in Section 3.9 of the text 2 .
Exercises Exercises
Study Calculus Volume 1, Section 3.9 3 ; in particular Examples 74, 75, 77, 78, 81 and 82, Checkpoint 54, and Exercises 333, 339, 347, 351 and 353.
We in particular emphasize the last topic of Logarithmic Differentiation, using the strategy of simplifying functions of the form \(\log(\dots)\) using the laws of logarithms like \(\log(a b) = \log(a) + \log(b)\text{.}\)
openstax.org/books/calculus-volume-1/pages/3-9-derivatives-of-exponential-and-logarithmic-functionsopenstax.org/books/calculus-volume-1/pages/3-9-derivatives-of-exponential-and-logarithmic-functionsopenstax.org/books/calculus-volume-1/pages/3-9-derivatives-of-exponential-and-logarithmic-functions
