if you find x2 and y2 very confusing, just do this: 1) Write the equation 2) Substitute f(x) with y 3) Substitute y with x and x with y 4) Solve for y 5) Replace y with f^-1(x) 6) Take the derivative 7) Finally, replace x with the given number Ex: 1) f(x) = x^3 + 3, [f^-1(11)]' 2) y = x^3 + 3 3) x = y^3 + 3 4) y = (3)square root of (x-3) 5) f^-1(x) = (3)square root of (x-3) f^-1(x) = (x-3)^(1/3) 6) f^(-1)'(x) = 1 / [3(x-3)^2/3] 7) f^(-1)'(x) = 1 / [3(11-3)^2/3] f^(-1)'(x) = 12
I love your channel. Oddly, i can't believe how much i miss its simplicity, its sentiments and the sheer intelligence it takes to understand it ..ur awesome
MR. Organic Chemistry Tutor, thank you for a solid analysis of the Derivatives of Inverse Functions in Calculus Two. This is one topic in Calculus that I do not understand from start to finish. I will rewatch this video and do problems in order to understand this material. This is an error free video/lecture on TH-cam TV with the Organic Chemistry Tutor.
You made the derivative of the inverse function much more complex than it really is. You were too technical for a struggling AP student to comprehend. I appreciate the various methods you applied to obtain an answer. Unfortunately, many students might become overwhelmed with the excess information that you displayed in your video. For the above-average AP students, your video is awesome. Yet for the average student, it might be too challenging to digest. To all AP students who need help with this concept view Alaina Vasta video. It is on TH-cam. I guarantee you that after watching her video you will have no problems solving this type of problem on the AP Calculus Exam.
Finding the derivative of the inverse has always been a struggle. Keep it simple, is my recommendation. The derivative of the inverse is 1 divided by the derivative of the function at the y value of the x value given. I would like to see more derivatives of inverses without the specific value, but just the function of the inverse's derivative.
At 5:14 you switched y and x , but did not solve for the inverse. I guess this means just switching the variables and doing implicit differentiation give you the equation for the slo P of the inverse?
8 minutes 40 seconds. You mention a leading coefficient test of the constant term and leading coefficient making the factors plus minus 1 and plus minus 5. Would factoring it to be (x+5)(x-1) = 0 be an incorrect method? I plugged in x =1 with this method and found it to be = 0.
I do not understand why he did not plug-in 3 after finding derivation of the function f'(3). Instead he found (x1,y1) (x2,y2) separately and plugged in 1.
that's the derivative of x^3 + 3, using the power rule. He's taking the derivative of the equation above. The little 1 looking symbol next to f is called prime, which means taking derivative of the function f.
5:21 HOW CAN THIS BE THE INVERSE FUNCTION when the inverse function is actually (x-3)^1/3 ???? You've only switched x and y, but haven't actually done any inversion ?!?!?!
@@karatekid8312 no it's not! Changing the variables doesn't mean it's the inverse, you actually have to swap it using algebra. I mean, what else does this supposed to mean: (x-3)^1/3 ? Just like the inverse of x^2 is x^1/2 RIGHT?
Derivatives - Formula Sheet: bit.ly/4dThzf1
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if you find x2 and y2 very confusing, just do this:
1) Write the equation
2) Substitute f(x) with y
3) Substitute y with x and x with y
4) Solve for y
5) Replace y with f^-1(x)
6) Take the derivative
7) Finally, replace x with the given number
Ex:
1) f(x) = x^3 + 3, [f^-1(11)]'
2) y = x^3 + 3
3) x = y^3 + 3
4) y = (3)square root of (x-3)
5) f^-1(x) = (3)square root of (x-3)
f^-1(x) = (x-3)^(1/3)
6) f^(-1)'(x) = 1 / [3(x-3)^2/3]
7) f^(-1)'(x) = 1 / [3(11-3)^2/3]
f^(-1)'(x) = 12
You made my Day Sir 🙂
Its 1/12
I love your channel. Oddly, i can't believe how much i miss its simplicity, its sentiments and the sheer intelligence it takes to understand it ..ur awesome
Finally... It clicked. This whole approach with tracking the X1 Y1, X2 Y2 saved my shit man. Hell yeah.
thx for your generous help I'm gonna cry...
Just remember its a wash out course lol. Keep that in mind.
@@cefb8923 whats that mean
@@Oschar157I’m guessing he means weed out, which means it’s a course that many people drop because it’s hard to
MR. Organic Chemistry Tutor, thank you for a solid analysis of the Derivatives of Inverse Functions in Calculus Two. This is one topic in Calculus that I do not understand from start to finish. I will rewatch this video and do problems in order to understand this material. This is an error free video/lecture on TH-cam TV with the Organic Chemistry Tutor.
You are literally getting me through my calculus course thank you so much!!!
my university professor couldn't explain this subject as well as you did, THANK YOU!
Best explanation in the web that I can find so far. Nice.
You made the derivative of the inverse function much more complex than it really is. You were too technical for a struggling AP student to comprehend. I appreciate the various methods you applied to obtain an answer. Unfortunately, many students might become overwhelmed with the excess information that you displayed in your video. For the above-average AP students, your video is awesome. Yet for the average student, it might be too challenging to digest. To all AP students who need help with this concept view Alaina Vasta video. It is on TH-cam. I guarantee you that after watching her video you will have no problems solving this type of problem on the AP Calculus Exam.
Finding the derivative of the inverse has always been a struggle. Keep it simple, is my recommendation. The derivative of the inverse is 1 divided by the derivative of the function at the y value of the x value given. I would like to see more derivatives of inverses without the specific value, but just the function of the inverse's derivative.
this one is a bit complicated but works like a charm.
Thank you very much, I love this VDO clip.
Amazing video, clear as the water of a swiss river in spring!
At 5:14 you switched y and x , but did not solve for the inverse. I guess this means just switching the variables and doing implicit differentiation give you the equation for the slo
P of the inverse?
You are an angel sent from heaven thank you
thank you
Helps a crap ton, thank you so much.
Thanks man u just saved a life from possibly academic doom ,if God wills it so I'll be lifting my certificate in ur honor
God bless this channel
Love u man👍🏻👍🏻👍🏻👍🏻👍🏻👍🏻👍🏻💋💋
8 minutes 40 seconds. You mention a leading coefficient test of the constant term and leading coefficient making the factors plus minus 1 and plus minus 5. Would factoring it to be (x+5)(x-1) = 0 be an incorrect method? I plugged in x =1 with this method and found it to be = 0.
Yo teach you dropped your crown 👑
gracias amigo
U helped me so muchhh thanks
I do not understand why he did not plug-in 3 after finding derivation of the function f'(3). Instead he found (x1,y1) (x2,y2) separately and plugged in 1.
Beautiful
This is so clear!
Finally, I found a person like me 🙂
How is f inverse of x equal to y2? 2:04
F(x2) = y1
F-1(x2) = y2
you are amazing
great video
thankk youuu
I’m totally gonna pass my exam on Friday 😏
❤
I literally can't keep track of all these numbers.... thats why i suck at inverses.
at 04:18 where did the 3x^2 come from? someone please help
that's the derivative of x^3 + 3, using the power rule. He's taking the derivative of the equation above. The little 1 looking symbol next to f is called prime, which means taking derivative of the function f.
+Harvir aha now I got it thank you.
The Derivative of a cubic function is a quadratic, and so on...
i love you
WHY DOES MY TEXTBOOK MAKE IT SOUND SO COMPLICATED THEN
no idea it always does
This is not a tutorial, it's you making exercises, we need more explanation!
stfu
Ur just stupid
He explains the process during the problems
5:21 HOW CAN THIS BE THE INVERSE FUNCTION when the inverse function is actually (x-3)^1/3 ???? You've only switched x and y, but haven't actually done any inversion ?!?!?!
it's literally the same thing!!
@@karatekid8312 no it's not! Changing the variables doesn't mean it's the inverse, you actually have to swap it using algebra. I mean, what else does this supposed to mean: (x-3)^1/3 ? Just like the inverse of x^2 is x^1/2 RIGHT?
I think you have a wrong answer it should be 1/24 not 1/12 just my opinion hahah because you forgot to times the 12 in 2
thank you.
❤