Logarithmic differentiation

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We use logarithms to help us differentiate.

Logarithms were originally developed as a computational tool. The key fact that made this possible is that:

\[ \log _b(xy) = \log _b(x)+\log _b(y). \]

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

\[ \left (1.38\cdot 10^{\answer [given]{2}}\right )\cdot \left (2.34 \cdot 10^{\answer [given]{1}}\right ). \]

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

\[ \log _{10}(1.38) \approx \answer [given]{0.1399}\qquad \text {and}\qquad \log _{10}(2.34)\approx \answer [given]{0.3692} \]

Add these numbers together to get $\answer [given]{0.5091}$. Essentially, we know the following at this point:

[Picture]

Using the table again, we see that $\log _{10}(\answer [given]{3.23})\approx 0.5091$. 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.

1 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

\[ \frac {d}{dx} \ln (f(x)) = \frac {1}{f(x)}\cdot f'(x) = \frac {f'(x)}{f(x)} \]

which follows from the chain rule. Let’s look at an illustrative example to see how this is actually used.

Compute:

\[ \frac {d}{dx} \frac {x^9e^{4x}}{\sqrt {x-4}} \]

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

\[ \frac {d}{dx} x^x \]

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 $f'(x)$,

\[ f'(x) = x^x + x^x\ln (x). \]

Compute the derivative.

\[ \frac {d}{dx}\ln {(|x|)} \]

The function $|x|$ is a piecewise defined function. If $x>0$ \begin{align*} \frac {d}{dx}\ln {(|x|)} &= \frac {d}{dx}\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*} \frac {d}{dx}\ln {(|x|)} &= \frac {d}{dx}\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.

\[ \frac {d}{dx}\ln {(|f(x)|)} \]

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

2 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 $n$,

\[ \frac {d}{dx} x^n = n x^{n-1}. \]

We will use logarithmic differentiation. Set $f(x) = x^n$. Write \begin{align*} \ln (|f(x)|) &= \ln \left (|x|^n\right ) , x\ne 0\\ &= 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.

2025-01-06 19:48:39