5:24 if confused at this point, to prove this is true, take a ln of both side (you get ln a= ln e ^lna), the rule of logs and exponents lets you bring down lna and put it as multiplier in front of lne. Since lne=1, you're left with lna= ln a, which is true. Therefore a is indeed = e to the power of (lna).
This is so well explained. After days and days of trying to get my head around a derivative of a forgetting curve this Mister Khan makes me see.... (don't know about tomorrow however, I will (according tot he equation I am working on - I will forget) but I suppose I could always come back). Thanks for your work Mister Khan.
Andrew6James ln(a) is a constant, that is why you are taking it as a constant. Cause you are taking the derivative with respect to X. If you were taking the derivative with respect to 'A' you would be right
totally agree with Carlos, that if you are derivating a constant in this case "ln(a)" which is a number, it'll be 0. Remember, ln(a) can be ln(2), ln(9999), or everything, but never be a variable ln(x)
Hi khanacademy,Your videos helped me so much but i have 1 question. Isn't khan academy a non profit organization. You might probably get money for having 2.6 million subs.
5:24 if confused at this point, to prove this is true, take a ln of both side (you get ln a= ln e ^lna), the rule of logs and exponents lets you bring down lna and put it as multiplier in front of lne. Since lne=1, you're left with lna= ln a, which is true. Therefore a is indeed = e to the power of (lna).
This is so well explained. After days and days of trying to get my head around a derivative of a forgetting curve this Mister Khan makes me see.... (don't know about tomorrow however, I will (according tot he equation I am working on - I will forget) but I suppose I could always come back). Thanks for your work Mister Khan.
kha you must be a saint.
Thanks sir you are the best teacher 😊🙏
thank you so much !
What functions make up the composite function: e^(ln(a)x) at [Time: 3:08]?
u(x) = e^x
v(x) = (ln a) x
composite function:
u(v(x)) = e^ ((ln a) x)
sorry I am a little confused, if the derivative of ln(x) is 1/x then why is the derivative of lna just lna? Should it not be 1/a?
Andrew6James ln(a) is a constant, that is why you are taking it as a constant. Cause you are taking the derivative with respect to X. If you were taking the derivative with respect to 'A' you would be right
totally agree with Carlos, that if you are derivating a constant in this case "ln(a)" which is a number, it'll be 0. Remember, ln(a) can be ln(2), ln(9999), or everything, but never be a variable ln(x)
@@carloscerda11 thanks a ton because I was searching for this since ages and was confused
:D
This almost stretches my math abilities to their limit 😉 but not quite. I'm glad that I basically get it, because that just about blew my mind.
You can also prove this using logarithmic differentiation.
y = a^x.
ln y = x·ln a.
1/y·dy/dx = ln a.
dy/dx = y· ln a.
dy/dx = a^x ln a. ◼
shouldn't a be a be a positive number greater than zero to make this work? Since you're using ln(a)
I guess: d/dx a^x and a < 0 = d/dx (-1) |a|^x. You can proceed then as shown in the video and e.g. d/dx -3^x will result in -3^x * ln(3)
Makes sense
t^2* e^-t=0.2, solve for t , how can we solve this
Well, I don't understand why (ln(a)*x)' = ln(a),. it's because of e^x'=e^x but just otherwise?
that's the point of the video?
Cute
Hi khanacademy,Your videos helped me so much but i have 1 question. Isn't khan academy a non profit organization. You might probably get money for having 2.6 million subs.
I think on TH-cam it's only possible to make money if there are ads on the video. This video has no ads...
Even if they do get money from ads, they would still use all that money on the website.
Maybe you're right?.
SquidwqrdPlayz remember non profits still have to pay salaries