derivative of sqrt(x+sqrt(x+sqrt(x+…))) - Videos

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derivative of sqrt(x+sqrt(x+sqrt(x+…))),
implicit differentiation,

blackpenredpen,
math for fun,

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28 COMMENTS

  1. Hmm so you say y = sqrt(x+sqrt(x+…)
    But you can rewrite this in infinitely many ways as this weird sum of square roots is going to infinity :
    y = sqrt(x+sqrt(x+…+y)
    So What is the real correct answer ? Nice video btw, glad to see more and more videos of maths on your channel these times!

  2. in these problems you always want to find the infinite sequence again, but somewhere else, then substitute.
    y = sqrt(x+sqrt(x+…
    if you take the square you get
    y²=x+ sqrt(x+sqrt(x+…
    then you can move the x to the left and you get
    y² – x = sqrt(x+sqrt(x+…
    From what we started with we know that y = sqrt(x+sqrt(x+…
    therefore:
    y² – x = y
    Then it's much simpler and you can use implicit differentiation 🙂

  3. I have a question. Is it not better to just express it in terms of x by solving the quadratic in the beginning? I had the same idea you had and solved it by Implicit Differentiation but then realized I could also do this:

    y^2-y-x=0
    D=1+4x

    Given that y is properly defined as a real number – albeit an irrational one maybe – it stands to reason that it is defined for all x:(1+4x>0). Moreover, since y=sqrt(x+y), if x<0, then y becomes the sqrt of something negative+the square root of something negative + …, which cannot be even a real number, let alone a positive one. Therefore, the domain for this function is (all x>0).

    Consequently, we solve the quadratic and obtain the following formula.

    y=1/2*(1+sqrt(1+4x)) , the negative-sqrt solution is considered invalid because:

    sqrt(1+4x)>1, for x>0
    1/2(1-sqrt(1+4x))<0, for x>0
    Therefore, y would be negative, which cannot be true given that y equals a square root of a positive real number.

    Now we can simply perform Explicit Differentiation onto the above formula and arrive at the derivative expressed in finitely many terms of x:

    dy/dx=2/sqrt(1+4x)

    What do you think?

    Edit: After taking a second look, the negative sqrt solution is also valid for 0>=x>=(-1/4)
    So, actually y is not a well defined function unless we specify the domain first.

    If x belongs between [-1/4,+inf), then y=1/2*(1+sqrt(1+4x))
    If x belongs between [-1/4,0], then y=1/2*(1-sqrt(1+4x))

    Either case, we can also use E.D.

  4. Am I crazy or that's the same than y = 1 + sqrt(1+4x)/2 and dy/dx = 1/sqrt(1+4x)???
    Here is how:
    y=sqrt(x+sqrt(x+sqrt(x+…. Squaring both sides…
    y^2=x+sqrt(x+(sqrt(x+…=x+y
    y^2-y-x=0 (quadratic equation with a=1, b=-1 and c=-x)
    y = (-b +/- sqrt(b^2-4ac))/2a
    y= (1 +/- sqrt(1+4x))/2
    Since x needs to non-negative (for how the function was defined first, wither you would have a sqrt of a negative number), the smallest result of the sqrt(1-4x) can be sqrt(1), and hence the subtraction in the numerator (if I use the the – option) would be negative for any x except zero, what would in turn would give a negative y, which is not possible with the original function if I take the sqrt sign as the principal root (the positive one). Hence I will discard the – option in the +/- and keep:
    y= (1+sqrt(1+4x))/2. which is easily differentiable:
    y=1/2+1/2*(1+4x)^1/2
    dy/dx= 1/2*1/2*(1+4x)^(-1/2)*4
    dy/dx= 1/sqrt(1+4x)

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