\(f(x) = (x^{2} + 3)^{2}\) is defines on the set of real numbers, R. Find the gradient of f(x) at x = \(\frac{1}{2}\).
A. 4.0 B. 6.5 C. 5.0 D. 10.6
Correct Answer: B
Explanation
\(f(x) = (x^{2} + 3)^{2}\) Using the chain rule, \(\frac{\mathrm d y}{\mathrm d x} = \frac{\mathrm d y}{\mathrm d u} \times \frac{\mathrm d u}{\mathrm d x}\) Let \(u = x^{2} + 3\) so that \(y = u^{2}\) \(\frac{\mathrm d y}{\mathrm d u} = 2u\) \(\frac{\mathrm d u}{\mathrm d x} = 2x\) \(\therefore \frac{\mathrm d y}{\mathrm d x} = 2u(2x) = 4xu\) But \(u = x^{2} + 3\), \(\frac{\mathrm d y}{\mathrm d x} = 4x(x^{2} + 3)\) At \(x = \frac{1}{2}, \frac{\mathrm d y}{\mathrm d x} = 4(\frac{1}{2})((\frac{1}{2})^{2} + 3)\) = \(2 \times \frac{13}{4} = \frac{13}{2} = 6.5\)