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third one . " [√(x+y) – √(x-y) ]²=1² (squared both side ). x+y+x-y-2√(x²-y²)=1 (let 1 to go left and let 2√(x²-y²) to go right 2x-1=2√(x²-y²) (square both side again) 4x²-4x+1=4x²-4y² make y² alone 4y²= 4x-1 2y=√(4x-1) [{{ at the beginin if x is pozitif y should be pozitif if y is negativ x+y <x-y and √(x+y) – √(x-y) cant be positive and 1 .if x is negative and if y is positive or negative √(x+y)-√(x-y) cant be realnumber and x ,yshould be positive }}] i know x and y is positive y=[√(4x-1)]÷2 now you can derivativ y , y'= 1/ √(4x-1)

Hey, i have an interesting question:- what is the derivative/integral of log base x of 2 (or any other number)

Awesome channel, isn't it? 🙂

#Yay

شكرًا لك

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قال رسول الله صلى الله عليه و سلم :

لايشكر الله من لا يشكر الناس

Gold as usual. I'd like to see you do some more actuarial probability problems. 😀

implicit method is better

I only like implicit differentiation for equations of higher degrees than 2, other than that, regular algebra seems much cleaner.

But I like your explanation for both, so #YAY for me.

Now I was wondering if you could make a video on complex based logarithms, as of a+b*i based logarithm of c+d*i in an e+f*i form. I know the Euler-formula would be great help in this.

in my opinion nicer is 2nd but with one diffrence :

when y^2=x-1/4 dont square both sides just diffrentiate and u get dy/dx*2y=1 divide by 2y and you have dy/dx=1/2y 😉

The 2nd way is better as there is less manipulation. Also, when you get to the stage of

x = y^2 + 0.25 you can do

dx/dy = 2y and then

dy/dx = 1/2y

I prefer implicit method, more fun.

There is no plus minus. Only the positive side. Redundant solutions might have been created by repeatedly squaring both sides of the equation.

https://www.desmos.com/calculator/jdnyblekhi

I usually like the first method but this time the second method gave us a function in terms of x, so it is better for me

You can actually find the value of y using the derivative and initial value.

y'=1/2y.

2ydy = dx.

y^2=x+c

Now we need to evaluate c, which happens to equal -1/4.

y=(-/+)sqrt(x-1/4)

: )

In the second method, instead of solving for y it seems like it would be easier to short-circuit at the step y^2 = x – 1/4, do implicit differentiation => 2y dy/dx = 1; immediately giving dy/dx = 1/(2y)

7:44 "It's not that bad"

I STRIKE AGAIN!!!

6:49 So why when we solve this differential equation (dy/dx=1/2y) we only get y = sqrt(x+c) and not this horrifying thing you've differentiated

No. 1500! How hard is it to hit a multiple of 10 in YouTube views?

Implicit differentiation should be the easiest.

√(x+y) – √(x–y) = 1 . . . . . . take differential of both sides

(dx + dy)/√(x+y) + (dy – dx)/√(x–y) = 0 . . . I've removed the common factor, ½; now isolate dy on the LHS, & dx on the RHS

[1/√(x+y) + 1/√(x–y)]dy = [1/√(x–y) – 1/√(x+y)]dx . . . Combine fractions and multiply through by √(x²–y²)

[√(x+y) + √(x–y)]dy = [√(x+y) – √(x–y)]dx . . . Now cross-divide to get:

dy/dx = [√(x+y) – √(x–y)] / [√(x+y) + √(x–y)] . . . Multiply top & bottom by that grand numerator, so as to 'conjugate' the denominator

= [x + y + x – y – 2√(x²–y²)] / [2y]

= [x – √(x²–y²)] / y

= x/y – √[(x/y)²–1]

After watching: Yes! That's a much better result! . . dy/dx = 1/(2y)

Incidentally, in the 2nd method, once you arrive at 4y² = 4x – 1, you know that you have a parabola; and that its axis of symmetry is parallel to the x-axis.

Fred

When you said that you were going to solve this in two different ways, I thought the 2nd way was going to be something like this: If we start with sqrt(x + y) – sqrt(x – y) = 1 then sqrt(x +y) = 1 + sqrt(x – y). Squaring both sides, x + y = 1 +2sqrt(x – y) + x – y. After some magic (algebra), 2y – 1= 2sqrt(x – y) => (2y – 1)^2 = 4(x -y). Then 2(2y-1)(2) = 4(dx/dy – 1). Simplifying 2y – 1 = dx/dy – 1 => dx/dy = 2y ∴ dy/dx = 1/2y ∎

long method…

3rd Method- Let y=xcos(2t). Simplification makes it easy.

Nice choice in problem and great video as always. Too bad you didn't realize the minus sign is not part of the solution. When you had 2y – 1 = 2sqrt(x-y), you could have noted this implies y > 1/2. Then it's easy to tell x > y from the original equation.

algebra first or dy/dx will be everywhere

I personally would do by implicit, but I must say the second way was much cleaner!

Uwow the left with implicit and the right with explicit, awesome

Sir please solve this

Frank and sofia investigate two positive integer A and B as follows:

1)frank picks a positive factor a of A

2) sofia picks a positive factor b of B and

3)they write down the product ab on sheet of paper

They repeat the above procedure for all possible order pairs (a,b).At the end,the calculate the of all numbers on the paper is 2340.if both A and B are divisible by 6 and have only 2 and 3 as their prime factors find the least possible value of A+B.

7:00 People became happy because the denominator was rationalised!

I sqear…. if this is your solution in your video, i quit, cuz i thought this was really clever:

find y' given sqrt(x+y)-sqrt(x-y)=1.

with a simple identity: sqrt(a)-sqrt(b)=sqrt(a-2sqrt(ab)+b)

sqrt(x+y-2sqrt(x^2+y^2)+x-y)=1

2x+2sqrt(x^2+y^2)=1

sqrt(x^2+y^2)=(.5-x)

y^2=(.5-x)^2 -x^2

y = sqrt((.5-x)^2 -x^2)

y'= .5((.5-x)^2-x^2)^-.5 *(-(1-2x) -2x)

y'= -.5/sqrt((.5-x)^2-x^2)