Question

If \(y = \sqrt{cosx^2+\sqrt{cosx^2+\sqrt{cosx^2+\ldots \infty }}}\) and \(\frac{dy}{dx} = \frac{f(x)}{2y-1}\) then, \(\int f(x)\,dx = \ldots\)

• A. \(sinx^2+c\)
• B. \(-sinx^2+c\)
• C. \(cosx^2+c\)
• D. \(-cosx^2+c\)

Hint:

Use the chain rule and the derivative of log((x+3)/x).

Answer 1

The Correct Option is C

Step 1: Chain rule:
Let \(u = \log\frac{x+3}{x} = \log(x + 3) - \log x\). Then \(\frac{d}{dx}\sin u = \cos u\cdot\frac{du}{dx}\).

Step 2: Derivative of u:
\[ \frac{du}{dx} = \frac{1}{x + 3} - \frac1x = \frac{x - (x + 3)}{x(x + 3)} = \frac{-3}{x(x+3)} \]

Step 3: Result:
\[ \frac{d}{dx}\sin\left(\log\frac{x+3}{x}\right) = \frac{-3}{x(x + 3)}\cos\left(\log\frac{x+3}{x}\right) \]
The negative sign is present because the ratio \(\frac{x+3}{x}\) decreases as x increases.

Final Answer:
The derivative is \(\frac{-3}{x(x+3)}\cos\left(\log\frac{x+3}{x}\right)\), option (C). \[ \boxed{\frac{-3}{x(x+3)}\cos\left(\log\frac{x+3}{x}\right)} \]