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Lecture 11: The Derivative of \( \ln x \)
GOAL: Differentiate \( \ln x \) and introduce the technique of Logarithmic Differentiation. Finish the proof of the Power Rule.
Summary:
\[ \begin{gathered} [\ln x]' = \frac{1}{x} \qquad [\log_b x]' = \frac{1}{x \ln b} \\[0.5em] [\ln f(x)]' = \frac{f'}{f} \qquad x^p = e^{p \ln x} \end{gathered} \]Example 1
Find \( y' \) if \( y = \dfrac{490t}{(t^2+7)^3} \)
\[ \begin{gathered} [490t]' = 490 \\[0.5em] \left[(t^2+7)^3\right]' = 3(t^2+7)^2 \cdot 2t = 6t\,(t^2+7)^2 \end{gathered}\] So, by the Quotient Rule, \[y' = \frac{490\,(t^2+7)^3 - 490t\,(6t)\,(t^2+7)^2}{(t^2+7)^6} \]
Now that we have access to all of the derivative rules, we can differentiate any
differentiable function, regardless of how complicated it is. This can get complicated quickly, so:
Goal Develop a tool to mitigate this
issue.
Derivative of \(ln (x)\)
Recall \( \left[e^{f(x)}\right]' = e^{f(x)} \cdot f'(x) \). We can use this to find \( \frac{d}{dx}(\ln x) \). We note that taking the derivative of both sides of an equation keeps in equality in tact.
\begin{align*} f(x) =&~ \ln x \\[0.5em] e^{f(x)} =&~ e^{\ln x}\\[0.5em] e^{f(x)} =&~ x \\[0.5em] \frac{d}{dx}\left(e^{f(x)}\right) =&~ \frac{d}{dx}(x) \\[0.5em] e^{f(x)} \cdot f'(x) =&~ 1 \\[0.5em] f'(x) =&~ \frac{1}{e^{f(x)}}\\[0.5em] f'(x)=&~\frac{1}{e^{\ln x}} \\[0.5em] f'(x)=&~ \frac{1}{x} \end{align*}Theorem
- \[ \frac{d}{dx}(\ln x) = \frac{1}{x} \]
- If \( f \) is a differentiable function: \[ \frac{d}{dx}\left[\ln(f(x))\right] = \frac{1}{f(x)} \cdot \left[f(x)\right]' = \frac{f'(x)}{f(x)} \]
Example 2
Find \( y' \) if \( y = \ln(x^2-1) \)
METHOD 1:
\[ y' = \left[\ln(x^2-1)\right]' = \frac{1}{x^2-1} \cdot 2x = \frac{2x}{x^2-1} \]METHOD 2:
\[ \begin{gathered} y = \ln(x^2-1) = \ln\left[(x-1)(x+1)\right] \\[0.5em] y = \ln(x-1) + \ln(x+1) \\[0.5em] y' = \frac{1}{x-1}(1) + \frac{1}{x+1}(1) \\[0.5em] = \frac{1}{x-1} + \frac{1}{x+1} \\[0.5em] = \frac{x+1 + x-1}{(x-1)(x+1)} = \frac{2x}{x^2-1} \end{gathered} \]So, in other words, we can use the Log Laws to break our function down into other functions that are easier to differentiate. Let's see another example.
Example 3
Find \( y' \) if \( y = \ln\left(\sqrt{\dfrac{x+1}{x^2+4}}\right) \)
\[ \begin{gathered} y = \ln\left(\left[\frac{x+1}{x^2+4}\right]^{\frac{1}{2}}\right) = \frac{1}{2}\ln\left(\left[\frac{x+1}{x^2+4}\right]\right) \\[0.5em] = \frac{1}{2}\left[\ln(x+1) - \ln(x^2+4)\right] \\[0.5em] y = \frac{1}{2}\ln(x+1) - \frac{1}{2}\ln(x^2+4) \\[0.5em] y' = \frac{1}{2} \cdot \frac{1}{x+1} \cdot 1 - \frac{1}{2} \cdot \frac{1}{x^2+4} \cdot 2x \\[0.5em] y' = \frac{1}{2x+2} - \frac{x}{x^2+4} \end{gathered} \]Example 4
Find \( y' \) if \( y = \log x = \log_{10} x \)
\[ \begin{gathered} y = \log_{10} x \xrightarrow{\text{Change of Base Formula}} \frac{\ln x}{\ln 10} \\[0.5em] y' = \frac{1}{\ln 10}\left[\ln x\right]' = \frac{1}{\ln 10} \cdot \frac{1}{x} = \frac{1}{x \ln 10} \end{gathered} \]In general, for a base \( b > 0 \),
\[ \frac{d}{dx}\left(\log_b x\right) = \frac{1}{x \ln b} \]Logarithmic Differentiation
Recall if \( y = f(x) \)
\[ \left[\ln y\right]' = \left[\ln(f(x))\right]' = \frac{1}{f(x)} \cdot f'(x) = \frac{y'}{y} \]We can use this to our advantage.
Example 5
Repeat Example 1 using this new technique. I.e., find \( y' \) if \( y = \dfrac{490t}{(t^2+7)^3} \)
If we take the natural log of both sides, we can simplify the right hand side using the Log Laws.
\begin{align*} \ln y =&~\ln\left(\frac{490t}{(t^2+7)^3}\right)\\[0.5em] \ln y =&~ \ln(490t) -\ln\left[(t^2+7)^3\right] \\[0.5em] \ln y=&~ \ln 490 + \ln t - \ln\left[(t^2+7)^{3}\right] \\[0.5em] \ln y =&~ \ln 490 + \ln t - 3\ln(t^2+7) \end{align*} Taking the derivative of both sides we get: \begin{align*} \frac{y'}{y} =&~ \frac{1}{t} - 3\cdot \frac{1}{t^2+7} \cdot 2t \\[0.5em] \frac{y'}{y} =&~\frac{1}{t} - \frac{6t}{t^2+7} \\[0.5em] y' =&~ y\left[\frac{1}{t} - \frac{6t}{t^2+7}\right] \end{align*} To complete the problem, we want \(y'\) to be solely a function of \(t\). I.e., replace \(y\) with \(\displaystyle \dfrac{490t}{(t^2+7)^3}\): \[ y' = \frac{490t}{(t^2+7)^3}\left[\frac{1}{t} - \frac{6t}{t^2+7}\right] \]Example 6
Find \( y' \) if \( y = (2x+1)^5(x^4-3)^6 \)
\begin{align*} \ln y =&~ \ln\left[(2x+1)^5(x^4-3)^6\right] \\[0.5em] \ln y =&~ \ln\left[(2x+1)^{5}\right] + \ln\left[(x^4-3)^{6}\right]\\[0.5em] \ln y =&~ 5\ln(2x+1) + 6\ln(x^4-3) \end{align*} Taking the derivative of both sides we obtain: \begin{align*} \frac{y'}{y} =&~ 5 \cdot \frac{1}{2x+1} \cdot 2 + 6\cdot \frac{1}{x^4-3} \cdot 4x^3 \\[0.5em] \frac{y'}{y} =&~ \frac{10}{2x+1} + \frac{24x^3}{x^4-3} \\[0.5em] y' =&~ y\left[\frac{10}{2x+1} + \frac{24x^3}{x^4-3}\right]\\[0.5em] y' =&~ (2x+1)^5(x^4-3)^6\left[\frac{10}{2x+1} +\frac{24x^3}{x^4-3}\right] \end{align*}Example 7
Find \( y' \) if \( y = x^x \)
\begin{align*} \ln y =&~ \ln\left(x^{x}\right)\\[0.5em] \ln y =&~ x \ln x \\[0.5em] \ln y =&~ x \ln x \end{align*} Taking the derivative of both sides we obtain: \begin{align*} \frac{y'}{y} =&~ 1 \cdot\ln x + x \cdot \frac{1}{x} \\[0.5em] \frac{y'}{y} =&~ \ln x +1 \\[0.5em] y' =&~ y\left[\ln x + 1\right] \\[0.5em] y' =&~ x^x\left[\ln x + 1\right] \end{align*}Example 8
If \( y = (x^2-8x+2)\sqrt{25-x^2} \), find \( y' \).
\begin{align*} \ln y =&~ \ln\left[(x^2-8x+2)\cdot\sqrt{25-x^2}\right] \\[0.5em] \ln y =&~ \ln\left(x^2-8x+2\right) + \ln\sqrt{25-x^2} \\[0.5em] \ln y =&~ \ln\left(x^2-8x+2\right) + \frac{1}{2}\ln\left(25-x^2\right) \end{align*} Taking the derivative of both sides we obtain: \begin{align*} \frac{y'}{y} =&~\frac{1}{x^2-8x+2}(2x-8) + \frac{1}{2}\cdot\frac{1}{25-x^2}(-2x) \\[0.5em] \frac{y'}{y}=&~ \frac{2x-8}{x^2-8x+2} - \frac{x}{25-x^2} \\[0.5em] y' =&~ y\left[\frac{2x-8}{x^2-8x+2} - \frac{x}{25-x^2}\right] \\[0.5em] y' =&~ (x^2-8x+2)\sqrt{25-x^2}\left[\frac{2x-8}{x^2-8x+2} - \frac{x}{25-x^2}\right] \end{align*}Example 9
Find \( y' \) if \( y = \dfrac{(x-2)^5\sqrt{x^2+1}}{(x+3)^7} \)
\begin{align*} \ln y =&~\ln\left[(x-2)^5\cdot\sqrt{x^2+1}\right] - \ln\left[(x+3)^7\right] \\[0.5em] \ln y =&~ \ln\left(x+2\right)^5 + \ln\left(\sqrt{x^2+1}\right) - \ln\left[(x+3)^7\right] \\[0.5em] \ln y =&~ 5\ln(x+2) + \frac{1}{2}\ln\left(x^2+1\right) - 7\ln(x+3) \end{align*} Taking the derivative of both sides we obtain: \begin{align*} \frac{y'}{y} =&~ 5\cdot\frac{1}{x+2} + \frac{1}{2}\cdot\frac{1}{x^2+1}\cdot 2x - 7\cdot\frac{1}{x+3} \\[0.5em] y' =&~ y\left[\frac{5}{x+2} + \frac{x}{x^2+1} - \frac{7}{x+3}\right] \\[0.5em] y' =&~ \frac{(x-2)^5\sqrt{x^2+1}}{(x+3)^7}\left[\frac{5}{x+2} + \frac{x}{x^2+1} - \frac{7}{x+3}\right] \end{align*}Derivative of \( x^p \)
If \( p\) is rational, then we saw the Power Rule holds in Lecture 10.
If \( p \) is irrational (e.g., \(\sqrt{2}, \pi, e\)), then \( x^p \) is only defined for \( x > 0 \). With this in mind,
\begin{align*} \left[x^p\right]' =&~\left[e^{\ln\left(x^{p}\right)}\right]' \\[0.5em] =&~ \left[e^{p\ln x}\right]' \\[0.5em] =&~ e^{p\ln x}\cdot p\cdot\frac{1}{x}\\[0.5em] =&~x^p\cdot p\cdot\frac{1}{x} \\[0.5em] =&~ p x^{p-1} \end{align*}