Local linearization | Derivative applications | Differential Calculus | Khan Academy

Differential calculus on Khan Academy: Limit introduction, squeeze theorem, and epsilon-delta definition of limits.

About Khan Academy: Khan Academy offers practice exercises, instructional videos, and a personalized learning dashboard that empower learners to study at their own pace in and outside of the classroom. We tackle math, science, computer programming, history, art history, economics, and more. Our math missions guide learners from kindergarten to calculus using state-of-the-art, adaptive technology that identifies strengths and learning gaps. We've also partnered with institutions like NASA, The Museum of Modern Art, The California Academy of Sciences, and MIT to offer specialized content.

For free. For everyone. Forever. #YouCanLearnAnything

Subscribe to KhanAcademy’s Differential Calculus channel:

[coolbrotherf]

This was a lot better than how my teacher explained it. I was so confused, but this helped a lot.

[Flamehara]

I've found that using point-slope form to find what you called L(X) made my final approximation a lot easier. After making an equation in this form (y-y1)=m(x-x1). Our y1 in this case was 2 and our x1 was 4. We find our m/slope with the good old power rule of the original equation to get 1/4. Our point-slope formula is now complete, reading Y-2=1/4(X-4). you simply plug in your new value (which in this case was 4.36) for X: Y-2=1/4(4.36-4). Solve for Y and you get 2.09!

[swordfern]

It's incredible that my teacher hands us spreadsheets and tells us to learn, while with this we're shown every step and are given an explanation on what's happening. Thanks!

[Didi-sgpf]

Thank you for helping me get a 100% on my calculus final!!!!

[tianhaosun]

I like how i got a math tutor ad on a khan academy video, I thought youtube had a good ad algorith. Khan for life.

[tylerianoakey]

Hi Professor, just wanted to say how much I enjoy your teaching abilities. Thanks for the fantastic videos.

[marshelglidden]

Hey Sal, just so you know, you can usually hold the ctrl key and use the scroll wheel to change the zoom on the current screen. Might be worth a shot later on.

[ziadhani]

Thanks for the videos. You helped a lot of people to understand things in Calculus.

[jeremy]

this is so much clearer than my math lesson at university !! <3

[haligali]

Man, you're amazing! I hope you're good in 2021!

[xit]

The way you came up with the line equation confused me a little. It was clearer to me when I put it in slope-intercept form:
(F(x) - F(4))/(x-4) = F'(4) and solving for F of x gives: F(x) = F'(4) (x-4) + F(4)

[veyselgonel]

Very clear lecture, thank you very much.

[syedrizvi]

THANK YOU! :D Khan Academy is such an amazing tool!

[Froggo_kek]

the guy who is talking is very talented at teaching

[sindrestokke]

Thank you so much, I wouldn't have understood this without this great video!

[FlashTrance]

Thanks for these videos, but fyi this one is out of order and very confusing in a couple parts. I didn't understand the derivative power rule at all (because we haven't yet learned it if we're watching in order) and was completely confused about how you "randomly" found out that f'(x) was 1/2x^-1/2.

I think the fact that you were using a re-arranged point-slope form for L(x) should also have been stated from the get-go. I ended up giving up on your equation for a while and found the same answer using mx+b and the value of 1/4 given for f'(4) (slope-intercept form is what was used in an earlier video as a method for finding a tangent line equation).

[VinsonTran]

Essentially the formula used is
f(a)+f'(a)(x-a) then plug and chug

[anilkumarp]

thank you sir for your help.and we are expecting more from you

[bushratarif]

Thankyou sir for giving us a great lecture.

[xit]

At 6:02 you said the change in x is 4.36. Shouldn't that be 0.36?