32): k(x)=x^2 sec ((1)/(x))

This appears to be the beginning of a calculus problem, likely asking for the derivative of the function k(x). Let's find that derivative.Given \( k(x) = x^2 \sec\left(\frac{1}{x}\right) \), we'll use the product rule and the chain rule to find k'(x).The product rule states: (fg)' = f'g + fg'Here, f(x) = x² and g(x) = sec(1/x)* f'(x) = 2x* To find g'(x), we use the chain rule: * The derivative of sec(u) is sec(u)tan(u) * du/dx * Let u = 1/x = x⁻¹ * du/dx = -x⁻² = -1/x² * So, g'(x) = sec(1/x)tan(1/x) * (-1/x²)Now, applying the product rule:k'(x) = (2x) * sec(1/x) + x² * [sec(1/x)tan(1/x) * (-1/x²)]k'(x) = 2x sec(1/x) - sec(1/x)tan(1/x)Therefore, the derivative of \( k(x)=x^{2} \sec \left(\frac{1}{x}\right) \) is \( 2x \sec\left(\frac{1}{x}\right) - \sec\left(\frac{1}{x}\right)\tan\left(\frac{1}{x}\right) \).