Lec 6 | MIT 18.01 Single Variable Calculus, Fall 2007
Why Mathematics Stack Exchange recommends it
Mathematics Stack Exchange askers cite it when they get stuck on a specific step in its derivation of e and exponential derivatives, referencing exact timestamps.
Recommended by 0 people across 0 different months, 2015 to 2020.
What it is
Lecture 6 of MIT 18.01 Single Variable Calculus (Fall 2007) covers exponential and logarithmic functions, logarithmic differentiation, and hyperbolic functions. It is a recorded classroom lecture from MIT OpenCourseWare, with the derivation of why the number e arises from the derivative of a^x.
What people said
Cites StanManTheMathPan · 2015-09 · Mathematics Stack Exchange · on “Calculus, specifically deriving the rule for exponents”
I was watching the MIT lectures series on calculus 1( [this video] ), at 29minutes and 19 seconds he converts a^x to base e by e^((log(a))^x) Why does he convert base a to base e by raising e to the power of the constant log(a)? Sorry if the answer is obvious, I've searched all over and can't seem to figure out why he did this.
Cites Aayush Agrawal · 2016-03 · Mathematics Stack Exchange · on “Why is the following limit operation valid?”
Towards the end of the lecture (Found here [this video] at 46:15), the following equality is stated: $\lim_{n \to \infty} (1 + \frac1n)^n = e^{\lim_{n \to \infty}ln((1+\frac1n)^n)}$ It was earlier proved that $lim_{n \to \infty}ln((1+\frac1n)^n) = 1$, so the above would then give a definition for $e$. However, i don't understand why the above statement is valid. By the definition of $e$ it's obvious that $e^{log(x)} = x$, however here the statement is instead $e^{\lim_{n \to something} log(somethingelse)}$.
Cites Yosi Pramajaya · 2020-06 · Mathematics Stack Exchange · on “Why $ \dfrac{d}{dx} f(kx) = kf'(kx) $?”
In MIT OCW 18.01 Single variable calculus lecture 6 [this video] , minute 18:00-18:20, the professor used an example to explain why $e$ exists. Context: $f(x) = 2^x$ $b=2^k$ $ f(kx)=2^{kx} = (2^k)^x $ Then the professor wrote this, that by chain rule: $$ \dfrac{d}{dx} f(kx) = kf'(kx) $$ Why does it work? Why need to add additional $k$ there? If I expand derivative of $f(x)$ as: $$ \dfrac{d}{dx}f(x) = 2^x\dfrac{(2^{\Delta x} - 1)}{\Delta x} $$ then derivative of $f(kx)$ should be: $$ \dfrac{d}{dx}f(kx) = f'(kx) = 2^{kx}\dfrac{(2^{\Delta x} - 1)}{\Delta x} $$ Even if it's typo where Professor intended to write: $\dfrac{d}{dx}f(kx) = kf'(x) $ I don't think that it will work either. Because $$ k 2^{x