derivative of sqrt(x+y)-sqrt(x-y)=1 (as seen on “so you think you can take the derivative”) - Videos

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31 COMMENTS

  1. 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)

  2. 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.

  3. 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)

  4. 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

  5. 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 ∎

  6. 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.

  7. 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.

  8. 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)

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