Find the derivatives of the functions in Exercises 1–42.


𝔂 = x⁻¹/² sec (2x)²

Verified step by step guidance

1. Identify the function to differentiate: \( y = x^{-1/2} \sec(2x)^2 \). This is a product of two functions: \( u = x^{-1/2} \) and \( v = \sec(2x)^2 \).
2. Apply the product rule for differentiation, which states that if \( y = uv \), then \( y' = u'v + uv' \).
3. Differentiate \( u = x^{-1/2} \) with respect to \( x \). Use the power rule: \( u' = -\frac{1}{2}x^{-3/2} \).
4. Differentiate \( v = \sec(2x)^2 \) with respect to \( x \). Use the chain rule: first differentiate \( \sec(2x) \) to get \( 2\sec(2x)\tan(2x) \), then apply the power rule to get \( v' = 2 \cdot 2\sec(2x)\tan(2x) \cdot \sec(2x) = 4\sec(2x)^2\tan(2x) \).
5. Substitute \( u' \), \( v \), \( u \), and \( v' \) into the product rule formula: \( y' = (-\frac{1}{2}x^{-3/2})\sec(2x)^2 + x^{-1/2}(4\sec(2x)^2\tan(2x)) \). Simplify the expression to find the derivative.