Question

If \[ f(x) = \sqrt{2x + \frac{4}{\sqrt{2x}}} \] then \[ f'(2) \] is equal to: 

• A. \( 0 \)
• B. \( -1 \)
• C. \( 1 \)
• D. \( 2 \)
• E. \( -2 \)

Hint:

When a function is of the form \( f(x) = u + \frac{k}{u} \), the derivative is \( f'(x) = u' (1 - \frac{k}{u^2}) \). At the point where \( u^2 = k \), the derivative becomes zero. Here, at \( x=2 \), \( (\sqrt{2x})^2 = 4 \), which matches the constant in the numerator.

Answer 1

The Correct Option is A

Concept: To find the derivative of a function involving square roots and fractions, we first simplify the expression using power notation \( \sqrt{u} = u^{1/2} \). We then apply the Power Rule \( \frac{d}{dx}x^n = nx^{n-1} \) and the Chain Rule where necessary. Finally, we substitute the given value of \( x \) into the resulting derivative function.

Step 1: Simplify and rewrite the function in power form.
The function is \( f(x) = \sqrt{2x} + \frac{4}{\sqrt{2x}} \). Let's rewrite this using exponents to make differentiation easier: \[ f(x) = (2x)^{1/2} + 4(2x)^{-1/2} \]

Step 2: Differentiate the function with respect to \( x \).
Applying the Power Rule and the Chain Rule (multiplying by the derivative of the inner function \( 2x \), which is \( 2 \)): \[ f'(x) = \frac{1}{2}(2x)^{-1/2} \cdot 2 + 4 \left( -\frac{1}{2} \right) (2x)^{-3/2} \cdot 2 \] \[ f'(x) = (2x)^{-1/2} - 4(2x)^{-3/2} \] Rewrite back into radical/fractional form: \[ f'(x) = \frac{1}{\sqrt{2x}} - \frac{4}{(\sqrt{2x})^3} \]

Step 3: Evaluate at \( x = 2 \).
Substitute \( x = 2 \) into the derivative: \[ f'(2) = \frac{1}{\sqrt{2(2)}} - \frac{4}{(\sqrt{2(2)})^3} \] \[ f'(2) = \frac{1}{\sqrt{4}} - \frac{4}{(\sqrt{4})^3} = \frac{1}{2} - \frac{4}{2^3} \] \[ f'(2) = \frac{1}{2} - \frac{4}{8} = \frac{1}{2} - \frac{1}{2} = 0 \]