For the second question you could just substitute cos(theta)sin(theta) with 1/2*sin(2*theta) first before differentiating so that it is slightly easier. Otherwise great video.
@@bprpcalculusbasicsah ok. All of the textbooks here teach the chain rule first before the product and quotient rule so I was a but suprised others do it the opposite way.
@@Ninja20704Hm, the AP Calculus textbook teaches the product and quotient rules first and then the chain rule. I think it’s because the terms product and quotient are familiar, whereas chain rule is a whole new concept
When I am teaching the product rule, I always put it in the same order as the numerator of the quotient rule, to make learning the quotient rule easier. When there are multiple factors going into the product rule, you can use a more general version: 1 term for each factor, taking the derivative of the factor corresponding to the term number. e.g. Let f(x) = g(x)h(x)j(x) f'(x) = g'(x)h(x)j(x) + g(x)h'(x)j(x) + g(x)h(x)j'(x) This can be continued ad nausium.
When I did calculus I literally never used the quotient rule like in the first question; I preferred to rewrite the function from [y = f/g] to [y = f(g^-1)] and then use the product rule, which I usually found to be simpler.
I could never remember the quotient rule so used f/g = f g^(-1) and the product rule. If you look at it, this is what the quotient rule gives: d/dx (f/g) = (g f' - f g')/(g^2) = f' / g - f g'/(g^2) = f' g^(-1) - f g' g^(-2) = f' g^(-1) + f (g^(-1))' = d/dx f g^(-1)
Semi-related: I myself am surprised to see when I see the quotient rule being taught before the chain rule as I learned the chain rule first, followed by the product rule and learned the quotient rule last, considering that the latter is essentially an extension of the product rule in combination of the chain rule.
this is the quotient rule in lyrical form I guess Not bad, but if you want my (admittedly unpopular) opinion, instead of remembering a mathematical rule you have to remember a line that helps you remember a mathematical rule. You're still remembering, why not remember what you want straight up? Anyway, whatever people want I suppose.
As always, thank you for making such a great video, but I think you can teach students to simiplify first before *blindly differentiating* things. For example, we can simiplify t/(1 + t) into (1 - 1/(1 + t)) first, such that we immediately know its derivative is 1/(1 + t)², as d/dx(1/x) = - 1/x². Then, applying product rule gives: d/dt( t sin t/(1 + t)) = d/dt[(1 - 1/(1 + t))sin t] = (sin t)/(1 + t)² + (cos t)(1 - 1/(1 + t)) = (sin t)/(1 + t)² + t(cos t)/(1 + t) = (sin t + t(1+t)(cos t))/(1 + t)² = (t cos t + sin t + t² cos t)/(1 + t)²
Part of the problem here (I feel) is the lack of teaching of manipulating fractions in earlier, school, maths along with the use of calculators. As it is so easy to type in a sum into a calculator, the ability to think and simplify to make the calculation to do mentally or by hand easier has been lost: students reach for a calculator _before_ thinking about what is being asked. This thought comes from my observation of the top maths class at a secondary school: when asked to divide by 1000 I can't remember why now, it may have been something like converting metres to kilometres) I observed most reaching for a calculator to do it. I was taught fraction manipulation at primary school before calculators were rife - the only calculating aids we had were log tables and analogue calculators (aka slide rules).
You could also mention the generalized product rule for more than two functions. In this case, the triple product rule would be: (fgh)’ = f’gh + fg’h + fgh’. In words, take the derivative of each part of the product one at a time and multiple by the “regular version” of the other parts.
Have you ever talked about the Weierstrass (universal or t) trig substitution? I recently learned it and found it extremely powerful, would be very cool if you made a video on it!
The second problem would probably have been simpler if we rewrote f using the double angle formula before differentiating. So f(x) = x(sin x)(cos x) becomes f(x) = (1/2)x(sin(2x)) and you only have to apply the product rule once.
Someone else pointed this and the answer they got from bprp was "True that. However I know that’s a HW question from James Stewart calculus book and it hasn’t covered the chain rule yet."
The main point of the derivative is to figure out the variation of the function, so i would advise to keep the products of cos(th)sin(th) and factor them. I understand you want to get to the same answer they found elsewhere thought.
Are we sure the questioner is human? AI might be further along than some people think. The 'ChatGPT tripping out' remark could be a ruse to avoid our suspicion.
Dude didn't even try to solve it most likely, and apparently can't use chatGPT either :D More seriously, this is a case of people not knowing how to ask questions. They wrote what they tried, but not what their issue was. The only actually helpful comment, which they didn't even take the time to reply to was "Your methods sound correct. Where are you getting stuck? Can you upload a photo of your work?"... I bet it's easier to have people do the problem for you right...
yeah. This looks like something I could do at school. I was always bad at math but differentiation was my favourite thing becouse it was so mechanicaly simple. You only have several rules and thats all. You don't need to think at all while doing it.
So ChatGPT only made a simple silly mistake on the first one, and got the second without converting to double angles. Not the worst performance. But still.....always check if you are using ChatGPT.
ChatGPT simply trains itself on the data provided to it, whether the data is right or wrong. If it got worse it's because other people can't do math, and ChatGPT copied them.
@@alanprak80 AI is simply linear regression in most cases. It predicts what word should come next. That means it's very good at sounding very confident, even when it's completely wrong.
For more derivative practice, check out 100 derivatives: ua-cam.com/video/AegzQ_dip8k/v-deo.htmlsi=qXligorCGU7jxFsf
2:00 and you have to remember the derivative of sin(t) [checks his wall chart]
I laughed.
Should be dy/dt in first problem. Not dy/dx.
Ah yes. Thanks for pointing that out.
I was waiting for him to pick up on it and correct it 😂
Me too@@AchtungBaby77
x is just t doing some push ups
bprp > ChatGPT
ChatBPRP
For the second question you could just substitute cos(theta)sin(theta) with 1/2*sin(2*theta) first before differentiating so that it is slightly easier. Otherwise great video.
True that. However I know that’s a HW question from James Stewart calculus book and it hasn’t covered the chain rule yet.
@@bprpcalculusbasicsah ok. All of the textbooks here teach the chain rule first before the product and quotient rule so I was a but suprised others do it the opposite way.
@@Ninja20704Hm, the AP Calculus textbook teaches the product and quotient rules first and then the chain rule. I think it’s because the terms product and quotient are familiar, whereas chain rule is a whole new concept
@@HeyKevinYT that is fair. For us we teach chain rule first because then we can at least prove the quotient rule using the chain and product.
@@Ninja20704are you indian??
When I am teaching the product rule, I always put it in the same order as the numerator of the quotient rule, to make learning the quotient rule easier.
When there are multiple factors going into the product rule, you can use a more general version: 1 term for each factor, taking the derivative of the factor corresponding to the term number. e.g.
Let f(x) = g(x)h(x)j(x)
f'(x) = g'(x)h(x)j(x) + g(x)h'(x)j(x) + g(x)h(x)j'(x)
This can be continued ad nausium.
When I did calculus I literally never used the quotient rule like in the first question; I preferred to rewrite the function from [y = f/g] to [y = f(g^-1)] and then use the product rule, which I usually found to be simpler.
especially since you don't have to remember which part is negative 🙂
Me too, always seemed an unnecessary thing to remember
I could never remember the quotient rule so used f/g = f g^(-1) and the product rule.
If you look at it, this is what the quotient rule gives:
d/dx (f/g) = (g f' - f g')/(g^2)
= f' / g - f g'/(g^2)
= f' g^(-1) - f g' g^(-2)
= f' g^(-1) + f (g^(-1))'
= d/dx f g^(-1)
Yes, I used calculus A LOT in my career (still do it for fun!), and always did it that way.
Semi-related: I myself am surprised to see when I see the quotient rule being taught before the chain rule as I learned the chain rule first, followed by the product rule and learned the quotient rule last, considering that the latter is essentially an extension of the product rule in combination of the chain rule.
low d-high minus high d-low
square the bottom and away we go
Eh?
@@xinpingdonohoe3978 kind of like the mnemonics for product and chain rule, helps beginners remember the formula
this is the quotient rule in lyrical form I guess
Not bad, but if you want my (admittedly unpopular) opinion, instead of remembering a mathematical rule you have to remember a line that helps you remember a mathematical rule. You're still remembering, why not remember what you want straight up? Anyway, whatever people want I suppose.
@@hi-ld4gg I feel like it's more trouble, confusion and corniness than it's worth.
@@xinpingdonohoe3978 varies by person though, and if it helps them remember it then I don't see why not
As always, thank you for making such a great video, but I think you can teach students to simiplify first before *blindly differentiating* things. For example, we can simiplify t/(1 + t) into (1 - 1/(1 + t)) first, such that we immediately know its derivative is 1/(1 + t)², as d/dx(1/x) = - 1/x². Then, applying product rule gives:
d/dt( t sin t/(1 + t))
= d/dt[(1 - 1/(1 + t))sin t]
= (sin t)/(1 + t)² + (cos t)(1 - 1/(1 + t))
= (sin t)/(1 + t)² + t(cos t)/(1 + t)
= (sin t + t(1+t)(cos t))/(1 + t)²
= (t cos t + sin t + t² cos t)/(1 + t)²
Part of the problem here (I feel) is the lack of teaching of manipulating fractions in earlier, school, maths along with the use of calculators.
As it is so easy to type in a sum into a calculator, the ability to think and simplify to make the calculation to do mentally or by hand easier has been lost: students reach for a calculator _before_ thinking about what is being asked.
This thought comes from my observation of the top maths class at a secondary school: when asked to divide by 1000 I can't remember why now, it may have been something like converting metres to kilometres) I observed most reaching for a calculator to do it.
I was taught fraction manipulation at primary school before calculators were rife - the only calculating aids we had were log tables and analogue calculators (aka slide rules).
You could also mention the generalized product rule for more than two functions. In this case, the triple product rule would be: (fgh)’ = f’gh + fg’h + fgh’. In words, take the derivative of each part of the product one at a time and multiple by the “regular version” of the other parts.
It's provable with induction using the two function rule.
Have you ever talked about the Weierstrass (universal or t) trig substitution? I recently learned it and found it extremely powerful, would be very cool if you made a video on it!
The second problem would probably have been simpler if we rewrote f using the double angle formula before differentiating.
So f(x) = x(sin x)(cos x) becomes f(x) = (1/2)x(sin(2x)) and you only have to apply the product rule once.
Someone else pointed this and the answer they got from bprp was "True that. However I know that’s a HW question from James Stewart calculus book and it hasn’t covered the chain rule yet."
dy/dt instead of dy/dx in 1st problem.
literally who cares
Thank you, sir
The main point of the derivative is to figure out the variation of the function, so i would advise to keep the products of cos(th)sin(th) and factor them. I understand you want to get to the same answer they found elsewhere thought.
okay did anyone seriously not notice all those boxes with red and black marker pens in the bottom right?
Whoever was using chatgpt for this is dumb.
Because wolfram alpha exists.
Make a video with derivatives of hyperbolic trig functions
Are we sure the questioner is human? AI might be further along than some people think. The 'ChatGPT tripping out' remark could be a ruse to avoid our suspicion.
Dude didn't even try to solve it most likely, and apparently can't use chatGPT either :D More seriously, this is a case of people not knowing how to ask questions. They wrote what they tried, but not what their issue was. The only actually helpful comment, which they didn't even take the time to reply to was "Your methods sound correct. Where are you getting stuck? Can you upload a photo of your work?"... I bet it's easier to have people do the problem for you right...
yeah. This looks like something I could do at school. I was always bad at math but differentiation was my favourite thing becouse it was so mechanicaly simple. You only have several rules and thats all. You don't need to think at all while doing it.
So ChatGPT only made a simple silly mistake on the first one, and got the second without converting to double angles. Not the worst performance. But still.....always check if you are using ChatGPT.
the thing about AI, is it is always very confident of its answers, caution always recommended
yeah its crazy how good 3.5 is
Am I the only one who thinks that ChatGPT has gone worse?
I kind of get what you mean. It's hard to place, but you might be on to something.
ChatGPT simply trains itself on the data provided to it, whether the data is right or wrong. If it got worse it's because other people can't do math, and ChatGPT copied them.
@@alanprak80 AI is simply linear regression in most cases. It predicts what word should come next. That means it's very good at sounding very confident, even when it's completely wrong.
Слухай чувачок, який вирішує офігенні проблеми: тільки dy/dt, а не dy/dx, бo dy/dx=0.
Should be dy/dt not dy/dx