Physics and Mathematics
Differentiation of Infinite Series
1. What is an Infinite Series?
An infinite series is the sum of infinitely many terms:
[y = a_{1} + a_{2} + a_{3} + a_{4} + \cdots]
If each term is a differentiable function of [x], then:
2. Rule for Differentiation of Infinite Series
Key Idea for All Nested-Radical Problems
If:
[y][ = \sqrt{f(x) + \sqrt{f(x) + \cdots}}]
Then rewrite:
[y][ = \sqrt{f(x) + y} \Rightarrow y^{2} = f(x) + y]
Differentiate ➝ factor ➝ isolate [\dfrac{dy}{dx}].
3. Examples (Step-by-Step)
(Nested Function Differentiation)
Example 1
[y][ = \sqrt{\sin x + \sqrt{\sin x + \sqrt{\sin x + \cdots}}}]
Let the repeating radical be [y]:
[y = \sqrt{\sin x + y}]
Square:
[y^{2} = \sin x + y]
Differentiate:
[2y \dfrac{dy}{dx} = \cos x + \dfrac{dy}{dx}]
Rearrange:
[(2y – 1)\dfrac{dy}{dx} = \cos x]
Final:
[\dfrac{dy}{dx} = \dfrac{\cos x}{2y – 1}]
Substitute original [y] if needed.
Example 2
[y][ = \sqrt{x + \sqrt{x + \sqrt{x + \cdots}}}]
Let [y] be the infinite expression:
[y = \sqrt{x + y}]
Square:
[y^{2} = x + y]
Differentiate:
[2y \dfrac{dy}{dx} = 1 + \dfrac{dy}{dx}]
Rearrange:
[(2y – 1)\dfrac{dy}{dx} = 1]
Final:
[\dfrac{dy}{dx} = \dfrac{1}{2y – 1}]
Example 3
[y][ = \sqrt{\tan x + \sqrt{\tan x + \sqrt{\tan x + \cdots}}}]
Let the repeated part = [y]:
[y = \sqrt{\tan x + y}]
Square:
[y^{2} = \tan x + y]
Differentiate:
[2y \dfrac{dy}{dx} = \sec^{2}x + \dfrac{dy}{dx}]
Rearrange:
[(2y – 1)\dfrac{dy}{dx} = \sec^{2}x]
Final:
[\dfrac{dy}{dx} = \dfrac{\sec^{2}x}{2y – 1}]
Example 4
[y][ = \sqrt{e^{x} + \sqrt{e^{x} + \sqrt{e^{x} + \cdots}}}]
Let [y] be the infinite radical:
[y = \sqrt{e^{x} + y}]
Square:
[y^{2} = e^{x} + y]
Differentiate:
[2y \dfrac{dy}{dx} = e^{x} + \dfrac{dy}{dx}]
Rearrange:
[(2y – 1)\dfrac{dy}{dx} = e^{x}]
Final:
[\dfrac{dy}{dx} = \dfrac{e^{x}}{2y – 1}]
Example 5
[y][ = \sqrt{\ln x + \sqrt{\ln x + \sqrt{\ln x + \cdots}}}]
Let:
[y = \sqrt{\ln x + y}]
Square:
[y^{2} = \ln x + y]
Differentiate:
[2y \dfrac{dy}{dx} = \dfrac{1}{x} + \dfrac{dy}{dx}]
Rearrange:
[(2y – 1)\dfrac{dy}{dx} = \dfrac{1}{x}]
Final:
[\dfrac{dy}{dx} = \dfrac{1}{x(2y – 1)}]
4. Conceptual Questions with Solutions
1. What is the main idea behind differentiating an infinite nested function?
We assume the infinite repeating part equals the function itself and rewrite it algebraically.
Example:
[y][ = \sqrt{f(x) + \sqrt{f(x) + \cdots}}]
⇒ [y = \sqrt{f(x) + y}]
⇒ Square and differentiate implicitly.
This reduces a complex infinite structure into a solvable equation.
2. Why do we square the equation like [y = \sqrt{f(x) + y}] before differentiating?
Because removing the square root gives:
[y^{2} = f(x) + y]
This form is easier to differentiate using algebra + implicit differentiation.
3. Does the differentiation always require implicit differentiation?
Yes.
Since [y] appears on both sides after simplification, we must differentiate with respect to [x] treating [y] as a function of [x].
4. What ensures that an infinite nested radical like [\sqrt{\sin x + \sqrt{\sin x + \cdots}}] actually converges?
The expression must produce a real and finite value.
General requirements:
[f(x) \ge 0]
The infinite sequence must converge (bounded + monotonic).
Example: [\sin x] must be non-negative → [x \in [0, \pi]].
5. In the expression [y^{2} = f(x) + y], why do we rearrange before differentiating?
Rearranging keeps derivative terms grouped:
[2y \dfrac{dy}{dx} = f'(x) + \dfrac{dy}{dx}]
⇒ [(2y – 1)\dfrac{dy}{dx} = f'(x)]
This allows solving for [\dfrac{dy}{dx}] directly.
6. Can we substitute the original infinite expression back for [y] after differentiation?
Yes — this yields a final answer in terms of elementary functions instead of the unknown variable [y].
Example:
If [y][ = \sqrt{x + \sqrt{x + \cdots}}], then [y] can be replaced by that entire radical in the final derivative.
7. What happens in differentiation if the infinite series does not converge?
Then the function is not well-defined → the derivative does not exist.
Differentiability requires a valid limit for the nested structure.
8. Why does the denominator often appear as [2y – 1] in the final derivative?
Because the algebraic step always leads to:
[y^{2} = f(x) + y]
⇒ Differentiate → [2y – 1] multiplies [\dfrac{dy}{dx}].
This pattern is characteristic of infinite radicals.
9. Do we always apply the chain rule while differentiating these expressions?
Yes.
Even though squaring simplifies the radical, [y] remains a function of [x], requiring the chain rule during implicit differentiation.
10. Why is defining the domain important before differentiating an infinite radical?
Because:
Nested square roots require the inside to be non-negative.
A derivative is meaningful only where the function is real and differentiable.
Example:
[y][ = \sqrt{\tan x + \sqrt{\tan x + \cdots}}]
⇒ Need [\tan x \ge 0] and denominator [2y – 1 \ne 0].
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