Calculus AB - The Chain Rule (Hard)
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- Опубликовано: 11 сен 2024
- This video lesson goes over three examples of using the chain rule where the algebra involved in finishing the problem can be rather challenging if your algebra skills are weak. Be patient... with practice these types of problems will become simple.
The negative power factoring used to really freak me out until i realized it works exactly the same way when factoring with natural powers. Thanks for the examples!
Hello good sir. it's so funny, earlier when I was doing my Calc hw I was complaining that I hated this f'n class. as I was watching your video though, it hit me!... I shouldn't be complaining because when I first started out in the most remedial of math (think elementary school math... fractions, decimals...) classes I remember saying I wanted to go to Calc and beyond. now that I'm here, I should want to excel, not mope and complain. seeing your video was an eye opener for me. it reminded me of how far I've come as a non trad student. thanks
HI can you please help me with this simutaneouse equation
x-y=6
xy=4
james willis
Get y by itself in terms of x in both equations, and you will get that y=4/x, and that y=x-6. Since both of those equations are now equal to y, you can set them equal to each other, and solve for x. You will have 4/x=(x-6)/1 (I am writing the second expression over 1, in order to think about it as a single rational expression). Cross-multiply to get x(x-6)=4. Distribute the x to get that x^2-6x=4. At that point, you have two options to solve for x:
1.) Complete the square, as the equation stands (You already have a coefficient of 1 on the x^2-term, as well as an even integer, as the coefficient of the x-term, so solving this quadratic equation by completing the square should be relatively nice, for this problem).
2.) Subtract 4 from both sides, to get that x^2-6x-4=0, and then use the Quadratic Formula, to solve for x (This quadratic equation is NOT factorable, since there are NO factors of -4 that add to -6) (In this case, a=1, b=-6, and c=-4). Remember that this is the Quadratic Formula:
x=((-b±sqrt(b^2-4ac))/(2a))
Option #1 is easier, provided that you have already learned how to complete the square. With Option #2, you would need to simplify the radical, and the expression completely, which is some extra steps that aren't really necessary.
Once you get your two answers for x, it is easiest to plug those values in for x, into the equation, y=x-6, in order to obtain the corresponding answers for y. You won't get any extraneous solutions, when you check your two solutions, by plugging them into the original equations, so both of your solutions will be valid.
I hope that helps.
It’s 3 am and my exam is at 8 am. Here I am lol we took this exam and everyone failed so he’s making us all retake it? Just some last min scramming. Also this is beautiful. Lol this comment itself motivated me I remember how smart I wanted to be when I grew up and now that I’m here it seems like I don’t wanna do the work.
nice explaination
#ycaguru
solid examples. we were presented with the chain rule in class, but not fully instructed on how to take the derivative of complex fractions. Thank you for the step-by-step instructions. I appreciate the parts where you explain that we are done with calculus
Thank you so much for this explanation! Feels like I have the worst math teacher ever, but this solved everything :)
Your handwriting is fantastic!
Hardest part is the algebra. Thanks for showing all work for biggest challenge.
omg! all of these examples are exactly in the order that i was doing my homework. Different problems i mean but same setup for each example after each other ahhahah thank you!
Who does homework
Thank you for making this! These exact problems stump me the most (as well as logarithmic and second derivatives)! Grrrrrr!
Thank you for this. I had to find the second derivative of (x squared + 6) to the 5/6. The factoring after using product rule when finding the second derivative stumped the hell out of me until I watched this video
This is really a very beneficial video. Understanding is made clear. Thanks
Why not use quotient rule in example 2?
In the second problem why dont we take out the x from 2x
Great explanations of hard examples
In the first example why is x^3 not in the final answer?
Isnt the fraction example soposed to be done with the quotient rule ???
Thanks so much for the vid! Helped me practice!
You are a good teacher.
Shoulden't the first answer be -3x^3-2x/(1-x^2)^1/2 or basically a negative answer
Superb video ! Thank you Jim.
Thanks, glad you liked it.
I'm at lost... where did the -1/2 from (1-x^2)^-1/2 come from ? I know that you pull the exponent to the front but why did he write it again ? and why negative?
savageGamer101 f(x)=x^n => f'(x)=nx^(n-1) It's the n-1.
@@AronHardemanGoogle Could you answer my question? For first problem where did the x^3 come from? Wouldn't that be integration rather than derivative?
@@glorymanheretosleep when he writes down the x^3, all the calculus is done already. The -x^3 comes from x^2 * 0.5 * -2x = -x^3 (simplified)
What does AP and AB abbreviate to?
And thanks for the lesson! It really helps me review for the final exam.
Advanced placement calculus is a kind of course you take in preparation for college afaik i'm european so i have no idea. The AB and BC are levels I and II respectively
Excelente colega
sir the explanation is so good .this made it easy for understanding....
No
Good Examples
This video saved my exam score
nice video very helpful
Great vid!
In example #2 why do you rewrite the cube root as x(x^2+4)^(-1/3)? Why is the power -1/3 and not simply 1/3?
it is because that factor was taken from the denominator, making it negative instead of positive.
thank you
Thank you!
thanks for teaching us
Hey my friend this is very easy
Mithat Gölbaşı it's east but not "very easy"
ABo NaBiL Gamer ! I am from Turkey.This exams very easy in Turkey
Fuck u!.
Do my home work genius!.
Thankyou very much sir
Why is it x^3 in the first example?
he multiplied x and x^2 thats why x^3.
Thanks a lot, Sir
You're welcome!
100th Like :)
God bless you
You did way more steps than you could’ve. DER of first term, leave second alone, then multiply DER of 1st by DER of second
thanks
it's like watching some playing jazz
In second step
How to come last term(-2x)
This is easy
HI can you please help me with this simutaneouse equation
x-y=6
xy=4
did u mean xy instead of xv??
because you cant have 3 variables in a simultaneous equation as far as I know
Substitute one equation into the other. x-y=6 so x=6+y so substitute that into other equation,
(6+y)y=4 so then y^2+6y-4=0
Solve the quadratic
x=-6.605551275464 and
x=0.60555127546399
Plug that back into one of the original equations to find the corresponding y and you have solved the system.
2 easy
I feel like im the only one who finds this really easy
The first one is wrong you cannot have a power of 1/2 in the denominator. You have to rationalize it.
Lol when you multiply the denominator by its conjugate it doesn’t make any difference but it makes the term simpler only because rationalization is considered a multiplication by one
I'm in 7th grade lmao. Just watching cause I'm bored as fuck
Thats hard !?!!?! Hahahahah
this vid is not for some "smart" people like you
you may find it simple (or pretend) but there are people who find it hard
Thank you
Thank you
Thank you