Logarithmic differentiation

$\newenvironment {prompt}{}{} \newcommand {\ungraded }[0]{} \newcommand {\todo }[0]{} \newcommand {\oiint }[0]{{\large \bigcirc }\kern -1.56em\iint } \newcommand {\mooculus }[0]{\textsf {\textbf {MOOC}\textnormal {\textsf {ULUS}}}} \newcommand {\npnoround }[0]{\nprounddigits {-1}} \newcommand {\npnoroundexp }[0]{\nproundexpdigits {-1}} \newcommand {\npunitcommand }[1]{\ensuremath {\mathrm {#1}}} \newcommand {\RR }[0]{\mathbb R} \newcommand {\R }[0]{\mathbb R} \newcommand {\N }[0]{\mathbb N} \newcommand {\Z }[0]{\mathbb Z} \newcommand {\sagemath }[0]{\textsf {SageMath}} \newcommand {\d }[0]{\mathop {}\!d} \newcommand {\l }[0]{\ell } \newcommand {\ddx }[0]{\frac {d}{\d x}} \newcommand {\zeroOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {0}{0}}}} \newcommand {\inftyOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {\infty }{\infty }}}} \newcommand {\zeroOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {0}{\infty }}}} \newcommand {\zeroTimesInfty }[0]{\ensuremath {\small \boldsymbol {0\cdot \infty }}} \newcommand {\inftyMinusInfty }[0]{\ensuremath {\small \boldsymbol {\infty -\infty }}} \newcommand {\oneToInfty }[0]{\ensuremath {\boldsymbol {1^\infty }}} \newcommand {\zeroToZero }[0]{\ensuremath {\boldsymbol {0^0}}} \newcommand {\inftyToZero }[0]{\ensuremath {\boldsymbol {\infty ^0}}} \newcommand {\numOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {\#}{0}}}} \newcommand {\dfn }[0]{\textbf } \newcommand {\unit }[0]{\mathop {}\!\mathrm } \newcommand {\eval }[1]{\bigg [ #1 \bigg ]} \newcommand {\seq }[1]{\left ( #1 \right )} \newcommand {\epsilon }[0]{\varepsilon } \newcommand {\phi }[0]{\varphi } \newcommand {\iff }[0]{\Leftrightarrow } \DeclareMathOperator {\arccot }{arccot} \DeclareMathOperator {\arcsec }{arcsec} \DeclareMathOperator {\arccsc }{arccsc} \DeclareMathOperator {\si }{Si} \DeclareMathOperator {\scal }{scal} \DeclareMathOperator {\sign }{sign} \newcommand {\arrowvec }[1]{{\overset {\rightharpoonup }{#1}}} \newcommand {\vec }[1]{{\overset {\boldsymbol {\rightharpoonup }}{\mathbf {#1}}}\hspace {0in}} \newcommand {\point }[1]{\left (#1\right )} \newcommand {\pt }[1]{\mathbf {#1}} \newcommand {\Lim }[2]{\lim _{\point {#1} \to \point {#2}}} \DeclareMathOperator {\proj }{\mathbf {proj}} \newcommand {\veci }[0]{{\boldsymbol {\hat {\imath }}}} \newcommand {\vecj }[0]{{\boldsymbol {\hat {\jmath }}}} \newcommand {\veck }[0]{{\boldsymbol {\hat {k}}}} \newcommand {\vecl }[0]{\vec {\boldsymbol {\l }}} \newcommand {\uvec }[1]{\mathbf {\hat {#1}}} \newcommand {\utan }[0]{\mathbf {\hat {t}}} \newcommand {\unormal }[0]{\mathbf {\hat {n}}} \newcommand {\ubinormal }[0]{\mathbf {\hat {b}}} \newcommand {\dotp }[0]{\bullet } \newcommand {\cross }[0]{\boldsymbol \times } \newcommand {\grad }[0]{\boldsymbol \nabla } \newcommand {\divergence }[0]{\grad \dotp } \newcommand {\curl }[0]{\grad \cross } \newcommand {\lto }[0]{\mathop {\longrightarrow \,}\limits } \newcommand {\bar }[0]{\overline } \newcommand {\surfaceColor }[0]{violet} \newcommand {\surfaceColorTwo }[0]{redyellow} \newcommand {\sliceColor }[0]{greenyellow} \newcommand {\vector }[1]{\left \langle #1\right \rangle } \newcommand {\sectionOutcomes }[0]{} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument }$

We use logarithms to help us differentiate.

Logarithms were originally developed as a computational tool, mainly due to their famous property: $\log _b(xy) = \log _b(x)+\log _b(y).$

Before the days of calculators and computers, this was critical knowledge for anyone in a computational discipline.

Compute $138\cdot 23.4$ using logarithms.

Start by writing both numbers in scientific notation $\left (1.38\cdot 10^{2}\right )\cdot \left (2.34 \cdot 10^{1}\right ).$ Next we use a log-table, which gives $\log _{10}(N)$ for values of $N$ ranging between $0$ and $9$ . We’ve reproduced part of such a table below.

From the table, we see that $\log _{10}(1.38) \approx 0.1399\qquad \text {and}\qquad \log _{10}(2.34)\approx 0.3692$ Add these numbers together to get $0.5091$ . Essentially, we know the following at this point:

Using the table again, we see that $\log _{10}(3.23)\approx 0.5091$ . Since we were working in scientific notation, we need to multiply this by $10^3$ . Our final answer is $3230 \approx 138\cdot 23.4$ Since $138\cdot 23.4 = 3229.2$ , this is a good approximation.

The moral is:

Logarithms allow us to use addition in place of multiplication.

Logarithmic differentiation

When taking derivatives, both the product rule and the quotient rule can be cumbersome to use. Logarithms will save the day. A key point is the following $\ddx \ln (f(x)) = \frac {1}{f(x)}\cdot f'(x) = \frac {f'(x)}{f(x)}$ which follows from the chain rule. Let’s look at an illustrative example to see how this is actually used.

Compute: $\ddx \frac {x^9e^{4x}}{\sqrt {x-4}}$

Recall the properties of logarithms:

$\log _b(xy) = \log _b(x) + \log _b(y)$
$\log _b(x/y) = \log _b(x) - \log _b(y)$
$\log _b(x^y) = y\log _b(x)$

While we could use the product and quotient rule to solve this problem, it would be tedious. Start by taking the logarithm of the function to be differentiated.

$\begin{align*} \ln\left(\frac{x^9e^{4x}}{\sqrt{x-4}} \right) &= \ln\left(\answer[given]{x^9e^{4x}}\right) - \ln\left(\answer[given]{\sqrt{x-4}}\right)\\ &= \ln\left(x^9\right)+\ln\left(e^{4x}\right) - \ln\left((x-4)^{1/2}\right)\\ &= \answer[given]{9}\ln(x)+4x - \answer[given]{\frac{1}{2}}\ln(x-4). \end{align*}$

Setting $f(x) = \frac {x^9e^{4x}}{\sqrt {x-4}}$ , we can write $\ln (f(x)) = 9\ln (x)+4x - \frac {1}{2}\ln (x-4).$ Differentiating both sides, we find $\frac {f'(x)}{f(x)} = \answer [given]{\frac {9}{x}+4} - \frac {1}{2(x-4)}.$ Finally we solve for $f'(x)$ , write $f'(x) = \left (\frac {9}{x}+4 - \frac {1}{2(x-4)}\right )\left (\answer [given]{\frac {x^9e^{4x}}{\sqrt {x-4}}}\right ).$

The process above is called logarithmic differentiation. Logarithmic differentiation allows us to compute new derivatives too.

For $x>0$ , compute $\ddx x^x$

The function $x^x$ is tricky to differentiate. We cannot use the power rule, as the exponent is not a constant; the function is not an exponential function either, since the base is not a constant. However, if we set $f(x) = x^x$ we can write $\begin{align*} \ln(f(x)) &= \ln\left(x^x\right)\\ &=x\ln(x). \end{align*}$

Differentiating both sides, we find $\frac {f'(x)}{f(x)} = \answer [given]{1 + \ln (x)}.$ Now we can solve for $f'(x)$ , $f'(x) = x^x + x^x\ln (x).$

Alternative way: We can write $x^x= e^{\ln {x^x}}=e^{x\ln {x}}.$ Therefore, $\ddx x^x=\ddx e^{x\ln {x}}=e^{x\ln {x}}\left (\ln {x}+1\right ),$ by the Chain Rule and the Product Rule.

Now, it follows that $\ddx x^x=x^x\left (\answer [given]{\ln {x}+1}\right ).$

Compute the derivative. $\ddx \ln {(|x|)}$

The function $|x|$ is a piecewise defined function. If $x>0$ $\begin{align*} \ddx\ln{(|x|)} &= \ddx\ln{(\answer[given]{x})}\\ &=\frac{1}{\answer[given]{x}}. \end{align*}$

The function $\ln {(|x|)}$ is not defined at $x=0$ . If $x<0$

$\begin{align*} \ddx\ln{(|x|)} &= \ddx\ln{(\answer[given]{-x})}\\ &=\frac{-1}{\answer[given]{-x}}\\ &=\answer[given]{\frac{1}{x}}. \end{align*}$

The next example will be useful when we want to use logarithmic differentiations for functions that assume negative values.

Compute the derivative. $\ddx \ln {(|f(x)|)}$

By the previous example and the Chain Rule $\begin{align*} \ddx\ln{(|f(x)|)} &=\answer[given]{\frac{1}{f(x)}}\cdot f'(x).\\ &=\frac{f'(x)}{\answer[given]{f(x)}} \end{align*}$

Proof of the power rule

Finally, recall that previously we only proved the power rule for positive integer exponents. Now we’ll use logarithmic differentiation to give a proof for all real-valued exponents. We restate the power rule for convenience sake:

Power Rule For any real number $n$ , $\ddx x^n = n x^{n-1}.$

We will use logarithmic differentiation. Set $f(x) = x^n$ . Write

$\begin{align*} \ln(|f(x)|) &= \ln\left(|x|^n\right) , x\ne0\\ &= n\ln(|x|). \end{align*}$

Now differentiate both sides, and solve for $f'(x)$

$\begin{align*} \frac{f'(x)}{f(x)} &= \frac{n}{x}\\ f'(x) &=\frac{n f(x)}{x}\\ &= \answer[given]{n x^{n-1}}. \end{align*}$

Whenever $x=0$ is in the domain of $f$ , the definition of the derivative at $x=0$ implies that $f'(0)=0$ , if $n\ne 1$ and $f'(0)=1$ , if $n=1$ , so the power rule applies to $x=0$ , too.

Thus we see that the power rule holds for all real-valued (nonzero) exponents.

While logarithmic differentiation might seem strange and new at first, with a little practice it will seem much more natural to you.