How do you find the derivative of sqrt(x+sqrt(x+sqrt(x+.......)))?
y = √(x+√(x+√(x+...)))
y^2 = x+√(x+√(x+...)))
y^2 = x+y
2yy' = 1+y'
y' = 1/(2y-1) = 1/[2√(x+√(x+...)))-1]
To find the derivative of the function √(x + √(x + √(x + ...))), we can consider it as an infinite sum of nested square roots. Let's denote this function as f(x).
To begin, let's notice that the function f(x) is essentially a limit. We can define f(x) recursively as follows:
f₀(x) = √(x)
fₙ₊₁(x) = √(x + fₙ(x))
Now, to find the derivative of this function, we need to apply some calculus techniques. Since f(x) is defined recursively, we will use the limit definition of the derivative.
First, let's calculate the derivative of the function fₙ₊₁(x) with respect to x:
fₙ₊₁'(x) = d/dx [√(x + fₙ(x))]
= 1/2(x + fₙ(x))⁻¹/² * (1 + fₙ'(x))
Next, let's find an expression for fₙ'(x). To do this, we can differentiate the recursive function fₙ(x):
f₀'(x) = (1/2)x⁻/²
Now, let's find fₙ'(x) for n > 0:
fₙ'(x) = d/dx [√(x + fₙ₋₁(x))]
= 1/2(x + fₙ₋₁(x))⁻¹/² * (1 + fₙ₋₁'(x))
Using this recursive definition, we can find the derivative of the function f(x).
By taking the limit as n approaches infinity, we can find the derivative of f(x) as:
f'(x) = lim(n->∞) fₙ'(x)
Evaluating this limit is a bit complicated, and it requires solving the recursive equation for fₙ'(x). Unfortunately, there is no simple, closed-form expression for the derivative of this function.
However, we can still use numerical approximation methods or computer programs to estimate the derivative at specific points or calculate its value within an interval.