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. Thanks for this, I am in calc BC and we have been doing logarithmic differentiation. I thought it would be fun to play with some infinite functions, like x^x^x…. What a coincidence!

3. Hi!, your videos are awesome!.
I would really appreciate if you help me with this interesting integral: x/(e^x -1) from 0 to infin. , the result is π^2/6. I don't know how to get that result.

Regards!

4. 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 🙂

5. 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.

6. Once you have y^2 – y = x, can't you just complete the square to solve for y? Then you'll have a closed form expression for y. Differentiate that, and get a much simpler answer for dy/dx.

7. 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)