2026 Fall Section 3.7 Derivatives of the Logarithmic Functions
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Overview
Ara Araeipour reviews logarithm rules and derives the natural-log derivative by implicit differentiation, then applies chain, product, and quotient rules to logarithmic expressions. The examples cover common and natural logs, change of base, composite arguments, and simplifications such as the derivative of one-half ln((1+x)/(1-x)) becoming 1/(1-x²).
Key takeaways
- Implicit differentiation of e^y = x gives the foundational rule d/dx[ln x] = 1/x, valid for x > 0.
- For a composite natural logarithm ln(g(x)), the chain rule produces g'(x)/g(x), so the argument's derivative must not be omitted.
- Expanding logarithms before differentiating can avoid unnecessary quotient-rule work; for example, ln(2x/(2x+1)) becomes ln 2 + ln x - ln(2x+1).
- A logarithm with base b other than e differentiates as g'(x)/(g(x) ln b), as shown by the derivative of log base 5 of x² + 1.
- Logarithm power rules can simplify composite expressions before differentiation: ln(x^3) = 3 ln x, so its derivative is 3/x.
- The derivative of one-half ln((1+x)/(1-x)) simplifies to 1/(1-x²) after applying the quotient property and combining the resulting fractions.
Chapters
- Reviews the power, product, quotient, and chain rules, including how to differentiate composite functions.
- Defines logarithms through the equivalence log base b of x equals y if and only if b to the y equals x.
- Recaps the logarithm product, quotient, and power rules, along with common logarithms in base 10 and natural logarithms in base e.
- Uses the change-of-base formula and the inverse relationship between y = e^x and y = ln x; ln x is negative for 0 < x < 1 and positive for x > 1.
- Evaluates log base 3 of 81 by rewriting 3^x = 81, giving x = 4.
- Converts log base 25 of 5 to 25^y = 5 and obtains y = 1/2.
- Uses powers of 10 to evaluate common logs, including log(100) = 2 and log(0.001) = -3.
- Rewrites logarithmic and natural-log equations as exponential equations, such as ln(s) = t becoming e^t = s.
- Starts from y = ln x and rewrites it as e^y = x.
- Differentiates implicitly to get e^y y' = 1, then substitutes e^y = x to derive y' = 1/x.
- Applies the chain rule to ln(g(x)), giving g'(x)/g(x) wherever the logarithm's argument is positive.
- Emphasizes checking the domain: ln x requires x > 0.
- Differentiates ln(5x - 1) by combining the reciprocal of the argument with the chain-rule factor 5.
- Shows two methods for ln(x^3): direct chain rule and the logarithm power rule, both yielding 3/x.
- Expands ln(2x/(2x+1)) into ln 2 + ln x - ln(2x+1), producing 1/[x(2x+1)].
- Uses log decomposition and product or quotient rules on expressions including ln(3x(x+4)^5), 5x ln(x²), and ln x / x.
- Applies the product rule to e^(2x) ln x, factoring out e^(2x) to obtain e^(2x)(2 ln x + 1/x).
- Uses log base conversion, log base b of x = ln x / ln b, to derive the derivative 1/(x ln b).
- Rewrites log base 3 of √x using √x = x^(1/2), then applies the power rule to get 1/(2x ln 3).
- Differentiates log base 5 of x² + 1 using the argument derivative 2x and the factor 1/ln 5, yielding 2x/((x²+1)ln 5).
- Differentiates ln(sin x) by multiplying 1/sin x by cos x, then recognizes the result as cot x.
- Works through y = ln(ln(5x)), applying the chain rule twice to obtain 1/[x ln(5x)].
- Keeps the non-e base factor ln 5 separate to avoid confusing it with part of the logarithm's argument.
- Begins with one-half ln((1+x)/(1-x)) and expands it as one-half times [ln(1+x) - ln(1-x)].
- Differentiates each natural logarithm by the reciprocal-argument rule, accounting for the negative derivative of 1 - x.
- Combines 1/(1+x) and 1/(1-x) over the common denominator (1+x)(1-x).
- Simplifies the result to 1/(1-x²).
Transcript excerpt
0:17 Good morning class. Here's chapter 3, derivatives. Section 3.7, derivatives of the logarithmic functions. So we've covered derivatives and rates of change, derivative as a function, basic differentiation formulas, product and quotient rules, the chain rule, implicit differentiation, related rates, and now logarithmic functions. And here's the synopsis of that. But let's remind everybody of
1:06 differentiation formulas. And it is extremely important to be comfortable with that because the reverse process anti-ifferiation also known as integration is coming up. So those are the definitions dy dx is limit of delta y over delta x of delta x approaches zero or frime of x this is in notation. This is frime of x limit of f of x plus hus f of x / h as h approaches zero. And if the derivative of x is one, if you have c * f of x, you can take the c out. The the derivative of a sum is sum of the derivatives and so on. Power rule, the derivative of x^ 1 is n x^ n minus one. As a result, the derivative of c is 0, kx is k. And these are all trigonometric functions. For example,
1:54 the derivative of sin x is cossine x and so on. uh product rule, quotient rule. Uh the derivative of fg prime is frime g plus g prime f. It's a shorthand notation and f g prime is frime gus g prime f over g. A composite function y= f of g of x results in a chain rule where y prime is f of g prime. That means f of frime of g of x * g prime of x. extended power rule when you have the derivative of g of x the^ of n is n g of x^ n - 1 g prime of
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Summary, takeaways, and chapters were generated by AI from the video's transcript and may contain errors. The video belongs to its creator, Professor Ara Araeipour.