12. [-/1 Points] DETAILS MY NOTES LARCALC12 2.4 .085. Find the second derivative of the lunction \[ \begin{array}{l} f(x)=\sin \left(2 x^{2}\right) \\ f(x)=1 \end{array} \] NeedHelp? Fand

To find the second derivative of the function \( f(x) = \sin(2x^2) \), we first differentiate it once. The first derivative \( f'(x) \) can be found using the chain rule: \[ f'(x) = \cos(2x^2) \cdot (4x) = 4x \cos(2x^2). \] Next, we differentiate \( f'(x) \) again to get the second derivative \( f''(x) \). Applying the product rule: \[ f''(x) = \frac{d}{dx}(4x) \cdot \cos(2x^2) + 4x \cdot \frac{d}{dx}(\cos(2x^2)). \] The second term requires the chain rule again. Simplifying gives us: \[ f''(x) = 4\cos(2x^2) - 16x^2\sin(2x^2). \] Now, let's bring some fun into it! Do you know that sine functions appear everywhere in oscillation and wave theory? From the gentle sway of a swing to the rhythms of music, sine waves encapsulate motion and sound in a harmonious dance. Next time you swing back and forth, you're witnessing the magic of sine in action! If you find yourself mixing up the rules of differentiation, don’t fret! Start by breaking down complex functions into simpler parts. Writing out the derivatives step-by-step and checking your work regularly can help prevent those sneaky mistakes. Practice makes perfect, and soon enough, you'll be differentiating with ease!