Question

If \(y = \sqrt{\cosh x + \sqrt{\cosh x}}\), then \(\frac{dy}{dx} =\)

• A. \(\frac{\sinh x (2y^2 + 2\cosh x + 1)}{4y(y^2 + \cosh x)} \)
• B. \(\frac{\sinh x (2y^2 - 2\cosh x - 1)}{4y(y^2 - \cosh x)} \)
• C. \(\frac{\sinh x(1-2\sqrt{\cosh x})}{4y\sqrt{\cosh x}} \)
• D. \(\frac{\sinh x(1+2\sqrt{\cosh x})}{4y\sqrt{\cosh x}} \)

Hint:

When differentiating composite functions involving nested square roots, it's often helpful to square the equation once or twice to simplify the expression before applying implicit differentiation. Remember the derivatives of hyperbolic functions: \(\frac{d}{dx}(\cosh x) = \sinh x\) and \(\frac{d}{dx}(\sinh x) = \cosh x\).

Answer 1

The Correct Option is D

Step 1: Simplify the given expression for \(y\).
The given function is \(y = \sqrt{\cosh x + \sqrt{\cosh x}}\).
To simplify differentiation, square both sides of the equation to remove the outermost square root: \[ y^2 = \cosh x + \sqrt{\cosh x} \] Step 2: Differentiate implicitly with respect to \(x\).
Differentiate both sides of the equation \(y^2 = \cosh x + \sqrt{\cosh x}\) with respect to \(x\).
For the Left Hand Side (LHS):
\(\frac{d}{dx}(y^2) = 2y \frac{dy}{dx}\) (using the chain rule). For the Right Hand Side (RHS):
Recall that \(\frac{d}{dx}(\cosh x) = \sinh x\).
For the term \(\sqrt{\cosh x}\), use the chain rule for \(\sqrt{u}\), where \(u = \cosh x\).
\(\frac{d}{dx}(\sqrt{\cosh x}) = \frac{1}{2\sqrt{\cosh x}} \cdot \frac{d}{dx}(\cosh x) = \frac{1}{2\sqrt{\cosh x}} \cdot \sinh x = \frac{\sinh x}{2\sqrt{\cosh x}}\).
Combining the derivatives of the terms on the RHS:
\(\frac{d}{dx}(\cosh x + \sqrt{\cosh x}) = \sinh x + \frac{\sinh x}{2\sqrt{\cosh x}}\).
Equating the derivatives of both sides: \[ 2y \frac{dy}{dx} = \sinh x + \frac{\sinh x}{2\sqrt{\cosh x}} \] Step 3: Factor out \(\sinh x\) and simplify the expression.
Factor out \(\sinh x\) from the terms on the RHS: \[ 2y \frac{dy}{dx} = \sinh x \left(1 + \frac{1}{2\sqrt{\cosh x}}\right) \] Combine the terms inside the parenthesis by finding a common denominator: \[ 2y \frac{dy}{dx} = \sinh x \left(\frac{2\sqrt{\cosh x}}{2\sqrt{\cosh x}} + \frac{1}{2\sqrt{\cosh x}}\right) \] \[ 2y \frac{dy}{dx} = \sinh x \left(\frac{2\sqrt{\cosh x} + 1}{2\sqrt{\cosh x}}\right) \] Step 4: Solve for \(\frac{dy}{dx}\).
Divide both sides by \(2y\): \[ \frac{dy}{dx} = \frac{\sinh x (2\sqrt{\cosh x} + 1)}{2y \cdot (2\sqrt{\cosh x})} \] \[ \frac{dy}{dx} = \frac{\sinh x (1 + 2\sqrt{\cosh x})}{4y\sqrt{\cosh x}} \] The final answer is $\boxed{\frac{\sinh x(1+2\sqrt{\cosh x})}{4y\sqrt{\cosh x}}}$.