Hardy Chavez
03/05/2024 · High School
\( \sqrt{x+\sqrt{x+\ldots}} t_{0} \propto \), find \( \frac{d y}{d x} \)
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Tutor-Verified Answer
Step-by-step Solution
To solve the equation \( y = \sqrt{x + \sqrt{x + \ldots}} \), we first need to express \( y \) in a more manageable form. We can set up the equation as follows:
\[
y = \sqrt{x + y}
\]
Next, we will square both sides to eliminate the square root:
\[
y^2 = x + y
\]
Rearranging this gives us a quadratic equation:
\[
y^2 - y - x = 0
\]
Now, we can use the quadratic formula to solve for \( y \):
\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
In our case, \( a = 1 \), \( b = -1 \), and \( c = -x \). Plugging these values into the formula gives:
\[
y = \frac{1 \pm \sqrt{1 + 4x}}{2}
\]
Since \( y \) must be non-negative (as it represents a square root), we take the positive root:
\[
y = \frac{1 + \sqrt{1 + 4x}}{2}
\]
Now, we want to find \( \frac{dy}{dx} \). We will differentiate \( y \) with respect to \( x \):
\[
\frac{dy}{dx} = \frac{d}{dx} \left( \frac{1 + \sqrt{1 + 4x}}{2} \right)
\]
Using the chain rule, we differentiate the square root:
\[
\frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{\sqrt{1 + 4x}} \cdot \frac{d}{dx}(1 + 4x)
\]
The derivative of \( 1 + 4x \) is \( 4 \), so we have:
\[
\frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{\sqrt{1 + 4x}} \cdot 4
\]
Simplifying this gives:
\[
\frac{dy}{dx} = \frac{2}{\sqrt{1 + 4x}}
\]
Thus, the final result is:
\[
\frac{dy}{dx} = \frac{2}{\sqrt{1 + 4x}}
\]
Quick Answer
\(\frac{dy}{dx} = \frac{2}{\sqrt{1 + 4x}}\)
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