Exponential functions differentiation intro | Advanced derivatives | AP Calculus AB | Khan Academy

[@kennymaccaferri2602]

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.

[@johnspence8141]

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).

[@tanishkchaturvedi8965]

Thanks sir you are the best teacher 😊🙏

[@SanjaySingh-rv8rw]

What functions make up the composite function: e^(ln(a)x) at [Time: 3:08]?

[@Daniel-iq3zv]

u(x) = e^x v(x) = (ln a) x composite function: u(v(x)) = e^ ((ln a) x)

[@Andrew6James]

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?

[@carloscerda11]

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

[@RioPradipto]

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)

[@manavgupta2059]

​@@carloscerda11 thanks a ton because I was searching for this since ages and was confused :D

[@cariboux2]

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.

[@sarahfarag9024]

thank you so much !

[@johnnolen8338]

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. ◼

[@sadpotato5111]

kha you must be a saint.

[@learnsanskrit_sabd]

t^2* e^-t=0.2, solve for t , how can we solve this

[@trollifjell1]

shouldn't a be a be a positive number greater than zero to make this work? Since you're using ln(a)

[@manuelk.8273]

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)

[@manavgupta2059]

Makes sense

[@sergey_zatsepin]

Well, I don't understand why (ln(a)*x)' = ln(a),. it's because of e^x'=e^x but just otherwise?

[@depayss2428]

that's the point of the video?

[@seandafny]

Cute

[@tsukasa6408]

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.

[@xD-jm2ie]

I think on KZbin it's only possible to make money if there are ads on the video. This video has no ads...

[@yazeed0ps3]

Even if they do get money from ads, they would still use all that money on the website.

[@tsukasa6408]

Maybe you're right?.

[@zoeredadams]

SquidwqrdPlayz remember non profits still have to pay salaries