Question

If $\sqrt{y} - \sqrt{y} - \dots = \sqrt{x} + \sqrt{x} + \dots$ then $dy/dx = \dots$

• A. $(y + x + 1)/(y - x + 1)$
• B. $(y - x - 1)/(y - x + 1)$
• C. $(y - x + 1)/(y - x - 1)$
• D. 1

Hint:

Converting an implicit equation involving infinite series into a set of parametric equations ($x=f(k)$ and $y=g(k)$) is often much faster and cleaner than attempting direct implicit differentiation.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given an equation equating two infinite nested radical series. We need to find the derivative $\frac{dy}{dx}$.

Step 2: Key Formula or Approach:
For an infinite repeating radical $L = \sqrt{a \pm \sqrt{a \pm \dots}}$, the entire nested term inside the primary root is identical to $L$ itself.
We can substitute $L$ back into the equation: $L = \sqrt{a \pm L}$.

Step 3: Detailed Explanation:
Let the common value of both infinite series be $k$.
Left Hand Side (LHS):
$$k = \sqrt{y - \sqrt{y - \dots}} \implies k = \sqrt{y - k}$$
Square both sides:
$$k^2 = y - k \implies y = k^2 + k$$
Right Hand Side (RHS):
$$k = \sqrt{x + \sqrt{x + \dots}} \implies k = \sqrt{x + k}$$
Square both sides:
$$k^2 = x + k \implies x = k^2 - k$$
We now have parametric equations for $y$ and $x$ in terms of the variable $k$.
Differentiate both with respect to $k$:
$$\frac{dy}{dk} = 2k + 1$$
$$\frac{dx}{dk} = 2k - 1$$
Using the chain rule to find $\frac{dy}{dx}$:
$$\frac{dy}{dx} = \frac{dy/dk}{dx/dk} = \frac{2k + 1}{2k - 1}$$
Now, we must express $k$ back in terms of $x$ and $y$.
Subtract the parametric equation for $x$ from $y$:
$$y - x = (k^2 + k) - (k^2 - k) = 2k$$
Substitute $2k = y - x$ directly into our derivative expression:
$$\frac{dy}{dx} = \frac{(y - x) + 1}{(y - x) - 1} = \frac{y - x + 1}{y - x - 1}$$

Step 4: Final Answer:
The derivative is $(y - x + 1)/(y - x - 1)$, which matches option (c).