@LakASambooodee Not sure. I am trying to vary the tutorials I put up across the modules to give everyone something. Eventually it will be done but in 2 months - not sure.
I know it's been said before, but thanks a lot. My class hasn't finished P3 (CIE maths) yet, and I'm using your vids as extra help/tutoring. Tell me, can P3 be completed in 2 months? Excluding past paper practice.
Hey, i think i have spotted a mistake, Surely the derivative of -5e^3x+2 will be -15e^3x+2. As you have said in some of your previous videos if y=e^ax+b then dy/dx = ae^ax+b... Look at 7.33 qs 3
Hi, I was just wondering with the chain rule, why do you always use the method of dy / dt x dt / dx. I thought the form dy / dx = n [ f (x) ] ^ (n-1) f ' (x) is a lot quicker and easier to do?
ExamSolutions hi can we write the first step as 6e^x by bringing the 2 to the front using the log rule of logb(mn) = n · logb(m).......your answer would be much appreciated!:D
No you cannot do that. You could say ln y = ln [3 e^(x squared)] which then gives ln y = ln(3) + ln[e^(x squared)] which is then ln y = ln(3) + x^2 You can then differentiate this implicitly but this is far too much work.
I actually got this answer myself. (I know it's been a year so you probably won't care or will' expectations figured this out or whatever, but it bares stating) when he differentiates 5e^-2x, he's doing it with respect to t, which is -2x, however, t has no coefficients and -2x is not in terms of t, so y = 5e^t when we differentiate with respect to t comes out as dy/dt= 5e^t which is the same as 5e^-2x, I hope this clarifies for anyone else who got this result ☺️ I hope my explanation is clear 😅
+glenn pedrosa hey i can help you with this. The dy/dx is always gonna equals to the sum of the intercept of the equivalent notation from the summation of the two factors of -0.5 and the constant change in the interation between the two.
{e^(-3x) + e^3x}^12. So i my math teacher did this on board back some months ago. He used Chain rule on this one. Here is where i got confused. d(e^(-3x)+e^3x)^12 / d(e^(-3x)+e^3x) and the result he wrote was 12(e^(-3x)+ e^3x). But i am thinking it should be 12(e^(-3x)+ e^3x)^11. Who is correct ?!
+ExamSolutions so I'm assuming I learnt the basics to calculus by anyway, I was told to work out the derivative of say y= 2x^3 + x + 7 you times the power by the front number and take 1 of the power. So dy/dx would become 6x^2 + 1.
+NortexG - Art That is correct but that is for very simple functions of the form ax^n. This is for more advanced functions and it seems that you have stumbled on this video which is further down the line from what you are learning.
This guy better be getting most out of UA-cam ad revenue, hes the only one who deserves it lol
A year later and im still using your tutorials to get me through my degree, i owe my good grades to you!
mate, I'm here a whole decade later lmao
Good to have your support still.
Good to hear. Time left for more revision then.
Pleasure. I hope my other videos continue to help
Thank you, thank you, thank you .....
Good to have your support, thanks.
Your videos have helped me so much. I cannot say thank you enough. Really.
THANK YOU SOOO MUCH!!!!!!!!
Very important video. He inspired me as well. Good job!
That is the rule for functions of the form ax^n not e^x
Your videos are put into a context which are really easy to understand!! thank you!
My guy i know you probably wont read this but if you do, YOU ARE SAVING MY LIFE. YOU ARE THE MESSIAH!!!
Well explained 👌
@LovelyAnnKay Most kind. Thanks
@LakASambooodee Not sure. I am trying to vary the tutorials I put up across the modules to give everyone something. Eventually it will be done but in 2 months - not sure.
i loved it very muchhh.. itz great.... tanxz very much Examsolutions....!!!!!!!!!! Im in debt to uuuu..... tanxz alot
great instructions, helps a lot!
really helpful video thank you so much
Thanks!
I know it's been said before, but thanks a lot. My class hasn't finished P3 (CIE maths) yet, and I'm using your vids as extra help/tutoring.
Tell me, can P3 be completed in 2 months? Excluding past paper practice.
Thanks
thanks alot
would you be able to do one on differentiation of exponential functions (not just e, like differentiating 2^x or sth) and logarithmic functions?
excellent!!!!!!!
@PinkyPurply Thanks for your support.
Hey, i think i have spotted a mistake, Surely the derivative of -5e^3x+2 will be -15e^3x+2. As you have said in some of your previous videos if y=e^ax+b then dy/dx = ae^ax+b... Look at 7.33 qs 3
Hi,
I was just wondering with the chain rule, why do you always use the method of dy / dt x dt / dx. I thought the form dy / dx = n [ f (x) ] ^ (n-1) f ' (x) is a lot quicker and easier to do?
ExamSolutions hi can we write the first step as 6e^x by bringing the 2 to the front using the log rule of logb(mn) = n · logb(m).......your answer would be much appreciated!:D
No you cannot do that.
You could say ln y = ln [3 e^(x squared)] which then gives
ln y = ln(3) + ln[e^(x squared)] which is then
ln y = ln(3) + x^2
You can then differentiate this implicitly but this is far too much work.
I know this video is the chain rule but can you differentiate using the quotient rule for question 4?
Please with kind I didn't understand this topic well can you repeat it please
For the last stage is acceptable to have (5e^t) where you have (5e^-2x), since t was defined to equal -2x?
Jay Agnihotri Not really. I would encourage you to sub it back in.
ExamSolutions
Ok thanks sir, will do.
Jay Agnihotri
(Did you mean within the working)?
Surely the answer to (4) is 20e^-2x as (5e^-2x) dy/dx = (-2)(5e^-2x)= -10e^-2x which, multiplied by -2 would give 20e^-2x?
Not at all. The answer is correct.
I actually got this answer myself. (I know it's been a year so you probably won't care or will' expectations figured this out or whatever, but it bares stating) when he differentiates 5e^-2x, he's doing it with respect to t, which is -2x, however, t has no coefficients and -2x is not in terms of t, so y = 5e^t when we differentiate with respect to t comes out as dy/dt= 5e^t which is the same as 5e^-2x, I hope this clarifies for anyone else who got this result ☺️ I hope my explanation is clear 😅
+Ronnie Kent *will've how did auto correct, correct to THAT?!
you keep on using e to the power of...
can you do equations with using e??
hi could please help me with these.given y=xe^-0.5 (x=summation of rational numbers).find dy/dx and show that y is equal to or less than k.please
+glenn pedrosa hey i can help you with this. The dy/dx is always gonna equals to the sum of the intercept of the equivalent notation from the summation of the two factors of -0.5 and the constant change in the interation between the two.
{e^(-3x) + e^3x}^12. So i my math teacher did this on board back some months ago. He used Chain rule on this one. Here is where i got confused. d(e^(-3x)+e^3x)^12 / d(e^(-3x)+e^3x) and the result he wrote was 12(e^(-3x)+ e^3x). But i am thinking it should be 12(e^(-3x)+ e^3x)^11. Who is correct ?!
It looks like you are both wrong according to what you have written. The answer is [ 12(e^(-3x)+ e^3x)^11 ] [ -3e^(-3x)+3e^3x]
+ExamSolutions Is this a combination of the power rule with the chain rule? do we continue with the multiplication process?
+C Richardson It is the chain rule and the differentials of e^ax where a is a constant.
nope still dont get it :( i was sure it would help with all the great ratings but nope not for this guy
hi i cannot seem to find an intergration videos
So you've just completely confused me... Not what I learnt it my lesson.
+NortexG - Art That's a shame. What did you learn in your lesson then. How does it differ?
+ExamSolutions so I'm assuming I learnt the basics to calculus by anyway, I was told to work out the derivative of say y= 2x^3 + x + 7 you times the power by the front number and take 1 of the power. So dy/dx would become 6x^2 + 1.
+NortexG - Art That is correct but that is for very simple functions of the form ax^n. This is for more advanced functions and it seems that you have stumbled on this video which is further down the line from what you are learning.
+ExamSolutions ah okay, I'm still learning at GCSE further maths. Thanks for the reply!
+NortexG - Art Okay. This is Advanced Level module C3 which is normally handled in year 13. Anyway. good luck with your course.
Making me want to fall a sleep,
get off the maths videos and go to videos on English.