I can't even begin telling you how these videos turned Calculus 1 around for me. My first test was a 60 which was the wake up call. Since then I've found these videos and haven't scored below a 90 on each exam.
I shall quit physical school and just watch Superman's videos. I hate math righ now, but when I understand it I love it. LOVE HATE relationship. Cal1 is not my friend and neither is cal 2, 3 and dif Q. I look forward to taking those classes -______-
This was so helpful. Thank you! My prof just rushes through things, never explains what or why he does things and just assumes we know how to simplify EVERYTHING. I wish he was this clear!
I think the most impressive thing here is how he memorized the mirrored function and uses those hand gestures to make it more clear for the students. If you think about it from his side its really impressive how hes so fluid with it. He has to flip the y axis on any graph he uses a hand gesture for.
at 4:21 function must be continuous at critical points (where f' is zero or not defined). Your rule is true only for differentiable functions. You restrict the class of functions.
If a function is differentiable, the function MUST be continuous. Existing critical points means that the function has a derivative or is differentiable. Eventually, you are not true.
for the function f(x) = 3x^(5/3) - 5x^(2/3): We determined in your video that the slope is undefined at x=0. However the original function is continuous at x=0. What does this mean?
If you look at the graph of the function, you can see that the slope is becoming vertical as we approach a sharp turn at x=0 . The slope of a vertical line is undefined, so that is why the first derivative is undefined at x=0 even though the function is continuous there. (The sharp turn also makes the derivative not exist at that point, this function just happens to have both a sharp turn and a vertical slope at the same point)
Starting around 17 minutes...I have a problem with him assigning a positive value to points with negative values under a radical. Seems like positive/imaginary for anything less than 0
@@actuary32174 idk if I misunderstood you or not but if you mean the radical in the second question its odd radical so it's legal to put negatives there
Try substituting -8. It's a cubed root not a square root, so the cubed root of -8 is -2. (Check that (-2)^3=-8). When you plug -8 into the numerator, you also get a negative answer. A negative numerator and a negative denominator give a positive final answer.
Functions are allowed to have sharp turns. The absolute value function f(x)=|x| is another example of a function that has a sharp turn at x=0. For the function in the lecture, there is a sharp turn at x=0 but the function is still continuous there. However if you look at the graph you can see that the slope gets steeper and steeper as we approach x=0. The slope becomes vertical, and the slope of a vertical line is undefined. Further, the slope is approaching positive infinity as we approach x=0 from the left, and the slope approaches negative infinity as we approach from the right. So even though the function is continuous at x=0, we see that the first derivative is undefined there so the function is not differentiable at x=0. That doesn't stop us from finding the derivative at the other points where it is defined.
here where Algebra is crutia and will not play with u! it you miss your critical numbers your answer will be in a critical condition and you will find yourself trying to catch up some points at the hospitals problems :D
I think so, yes. I studied pretty much this exact content when I studied "Differential Calculus" on my first year in Engineering Physics. Then my final math course that year was "Multivariable Calculus", which is basically Calculus 3.
f(x)=4x+2/2x-3 (4x plus 2, over 2x minus 3). The (1st derivatives) f'(x)= -16/(2x-3)^2 (negative 16 over 2x minus 3 raise to the power of 2), and the f"(x)= 64/(2x-3)^3 When ask when is f'(x) is increasing or decreasing. How to compute when the numerator has no x value). In (second derivatives) f"(x), how to compute when f(x) is positive(conveks) or when it is negative (concave)? Can someone at least explain the logic behind it? Eventually youtube link if there is any.
Man is both mentally and physically strong. That's someone to look up to. Great lesson and explanation
"no.. oh no, I'd burn your paper for that... hand ya back a staple!"
This dude is too much hahaha
This guys amazing. Wherever he is teaching, they better be paying him good !
Jbh
I can't even begin telling you how these videos turned Calculus 1 around for me. My first test was a 60 which was the wake up call. Since then I've found these videos and haven't scored below a 90 on each exam.
I shall quit physical school and just watch Superman's videos. I hate math righ now, but when I understand it I love it. LOVE HATE relationship. Cal1 is not my friend and neither is cal 2, 3 and dif Q. I look forward to taking those classes -______-
This was so helpful. Thank you! My prof just rushes through things, never explains what or why he does things and just assumes we know how to simplify EVERYTHING. I wish he was this clear!
Professor Leonard is really out here doing God's work
I think the most impressive thing here is how he memorized the mirrored function and uses those hand gestures to make it more clear for the students. If you think about it from his side its really impressive how hes so fluid with it. He has to flip the y axis on any graph he uses a hand gesture for.
Wish my professor was this clear helped a lot thanks
+SimplyGrouch Same
Stop being a weeb, things will become much clearer.
@@SlazeM7 HAHAHAHAHAHAHAA
Nobody talking abt Prof.'s humour👑
Thank you for going the extra step and posting these videos. I really appreciate your style and approach, it is very refreshing.
Best of luck
Casey
I wish you were my actual teacher but still I am thankful to god to have access to this
This awesome. Teaches better than my actual professor. Thanks
your videos are very straight forward . thank you for sharing the videos .
He's simply the best.
1:15 I'll burn your paper and hand you back the staple... LOL
Super helpful for my Ap Calc class, thanks
thank you for uploading these videos they help a lot
Hor thing and lesson was good :)Thanks superprofessor
wish I found this professor earlier. My current one is just making calculus way more complicated than this.
ممتاز هالدكتور
this man is a goddamn american hero
i took x^(2/3) as a factor and come up with the same answer on 18:33
@21:42 if x = 0 doesn’t it mean that graph has a vertical tangent ?
yes and it can also possibly change signs (slope from decreasing to increasing and vise versa) there
THANK YOU PROFESSOR!
Professor Leonard drop the workout split
so great!
at 4:21 function must be continuous at critical points (where f' is zero or not defined). Your rule is true only for differentiable functions. You restrict the class of functions.
If a function is differentiable, the function MUST be continuous. Existing critical points means that the function has a derivative or is differentiable. Eventually, you are not true.
But at 22:39 cube Root of x should not be equal to zero.
how did you pull out X ^-1/3 from X^2/3 and are left with an x?
Thank you! that really helped.
hey, i wanted u to do deltan and epsilon form limit defination,
please
I wish my teacher was also jacked maybe i would have focused more
for the function f(x) = 3x^(5/3) - 5x^(2/3): We determined in your video that the slope is undefined at x=0. However the original function is continuous at x=0. What does this mean?
If you look at the graph of the function, you can see that the slope is becoming vertical as we approach a sharp turn at x=0 . The slope of a vertical line is undefined, so that is why the first derivative is undefined at x=0 even though the function is continuous there. (The sharp turn also makes the derivative not exist at that point, this function just happens to have both a sharp turn and a vertical slope at the same point)
@ 9:23 Professor why are you flipping me off? lol
It's lucky for not having to take this test in the class :D
Starting around 17 minutes...I have a problem with him assigning a positive value to points with negative values under a radical. Seems like positive/imaginary for anything less than 0
This concerns me too. I agree with your point.
@@actuary32174 idk if I misunderstood you or not but if you mean the radical in the second question its odd radical so it's legal to put negatives there
@@moratasa9741ah that makes sense. Thanks
Try substituting -8. It's a cubed root not a square root, so the cubed root of -8 is -2. (Check that (-2)^3=-8). When you plug -8 into the numerator, you also get a negative answer. A negative numerator and a negative denominator give a positive final answer.
How can f(x) = 3x^(5/3) - 15x^(2/3) be a function when it has sharp turns at f(0?
Functions are allowed to have sharp turns. The absolute value function f(x)=|x| is another example of a function that has a sharp turn at x=0. For the function in the lecture, there is a sharp turn at x=0 but the function is still continuous there. However if you look at the graph you can see that the slope gets steeper and steeper as we approach x=0. The slope becomes vertical, and the slope of a vertical line is undefined. Further, the slope is approaching positive infinity as we approach x=0 from the left, and the slope approaches negative infinity as we approach from the right. So even though the function is continuous at x=0, we see that the first derivative is undefined there so the function is not differentiable at x=0. That doesn't stop us from finding the derivative at the other points where it is defined.
May I have the ISBN code for the calc text you are using.
thank youuu
How did you get the x^-1/3? How would 5^-1/3 * x = 5^2/3s?
Because x has an exponent of 1 therefore when factored it would be 1+-1/3=2/3 leaving it with 5^2/3
what edition is this Prof. Leonard? because i have edition 10th in my calculus and the chapters are different
bro why's that one kid always coughing in the back
here where Algebra is crutia and will not play with u! it you miss your critical numbers your answer will be in a critical condition and you will find yourself trying to catch up some points at the hospitals problems :D
Stupid question but is this the explanation for section 3.3 James Stewart 8th edition "how derivatives affect the slope of the graph"
Yes
my math teacher should kiss your shoes
Are these videos for first year university calculus?
I think so, yes.
I studied pretty much this exact content when I studied "Differential Calculus" on my first year in Engineering Physics.
Then my final math course that year was "Multivariable Calculus", which is basically Calculus 3.
f(x)=4x+2/2x-3 (4x plus 2, over 2x minus 3). The (1st derivatives) f'(x)= -16/(2x-3)^2 (negative 16 over 2x minus 3 raise to the power of 2), and the f"(x)= 64/(2x-3)^3 When ask when is f'(x) is increasing or decreasing. How to compute when the numerator has no x value). In (second derivatives) f"(x), how to compute when f(x) is positive(conveks) or when it is negative (concave)? Can someone at least explain the logic behind it? Eventually youtube link if there is any.
My goat
My fiancee also hates math.
Bro forget the calculus how can I have his physique 💀
ينطي واهس يقرا 😂
16:10 am i the only one heard the fly?
2 to 3
I would die for you
He dabs at 24:37
bruh wear a full shirt, your biceps are too distracting(in a good way(like a compliment, but like, bad for me))
lol @ how to find hors
+kinetic static go to the nearest hooker place
lolllll