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12. [0/1 Points] DETAILS MYNOTES LARCALC 122.4 .085 P Find the second derivative of the function: \[ \begin{array}{c} f(x)=\sin \left(8 x^{2}\right) \\ f^{\prime}(x)=4 \cos \left(2 x^{2}\right)=16 x^{2} \sin \left(2 x^{2}\right) \end{array} \] Need Holp? \( \square \) Read It

Ask by Martin Campos. in the United States
Feb 26,2025

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The second derivative of \( f(x) = \sin(8x^2) \) is \( f''(x) = 16\cos(8x^2) - 256x^2\sin(8x^2) \).

Solución

Find the second derivative of the function \( f(x)=\sin(8x^2) \). Evaluate the derivative by following steps: - step0: Evaluate the derivative: \(f^{\prime}\left(x\right)=\frac{d}{dx}\left(\sin\left(8x^{2}\right)\right)\) - step1: Use differentiation rules: \(f^{\prime}\left(x\right)=\frac{d}{dg}\left(\sin\left(g\right)\right)\times \frac{d}{dx}\left(8x^{2}\right)\) - step2: Find the derivative: \(f^{\prime}\left(x\right)=\cos\left(g\right)\times \frac{d}{dx}\left(8x^{2}\right)\) - step3: Calculate: \(f^{\prime}\left(x\right)=\cos\left(g\right)\times 16x\) - step4: Substitute back: \(f^{\prime}\left(x\right)=\cos\left(8x^{2}\right)\times 16x\) - step5: Multiply the terms: \(f^{\prime}\left(x\right)=16x\cos\left(8x^{2}\right)\) - step6: Evaluate: \(f^{\prime\prime}\left(x\right)=\frac{d}{dx}\left(16x\cos\left(8x^{2}\right)\right)\) - step7: Use differentiation rules: \(f^{\prime\prime}\left(x\right)=\frac{d}{dx}\left(16\right)\times x\cos\left(8x^{2}\right)+16\times \frac{d}{dx}\left(x\right)\times \cos\left(8x^{2}\right)+16x\times \frac{d}{dx}\left(\cos\left(8x^{2}\right)\right)\) - step8: Find the derivative: \(f^{\prime\prime}\left(x\right)=0\times x\cos\left(8x^{2}\right)+16\times \frac{d}{dx}\left(x\right)\times \cos\left(8x^{2}\right)+16x\times \frac{d}{dx}\left(\cos\left(8x^{2}\right)\right)\) - step9: Calculate: \(f^{\prime\prime}\left(x\right)=0+16\times \frac{d}{dx}\left(x\right)\times \cos\left(8x^{2}\right)+16x\times \frac{d}{dx}\left(\cos\left(8x^{2}\right)\right)\) - step10: Calculate: \(f^{\prime\prime}\left(x\right)=0+16\times 1\times \cos\left(8x^{2}\right)+16x\times \frac{d}{dx}\left(\cos\left(8x^{2}\right)\right)\) - step11: Calculate: \(f^{\prime\prime}\left(x\right)=0+16\cos\left(8x^{2}\right)+16x\times \frac{d}{dx}\left(\cos\left(8x^{2}\right)\right)\) - step12: Calculate: \(f^{\prime\prime}\left(x\right)=0+16\cos\left(8x^{2}\right)+16x\left(-16x\sin\left(8x^{2}\right)\right)\) - step13: Calculate: \(f^{\prime\prime}\left(x\right)=0+16\cos\left(8x^{2}\right)-16^{2}x^{2}\sin\left(8x^{2}\right)\) - step14: Calculate: \(f^{\prime\prime}\left(x\right)=16\cos\left(8x^{2}\right)-256x^{2}\sin\left(8x^{2}\right)\) The second derivative of the function \( f(x) = \sin(8x^2) \) is \( f''(x) = 16\cos(8x^2) - 256x^2\sin(8x^2) \).

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To find the second derivative of the function \( f(x) = \sin(8x^2) \), we first correctly differentiate \( f'(x) \). The first derivative you've noted seems to involve a mistake. Starting from \( f'(x) = \frac{d}{dx}[\sin(8x^2)] = 16x \cos(8x^2) \). Now for the second derivative, we need to differentiate \( f'(x) = 16x \cos(8x^2) \). Applying the product rule here will lead us to the correct second derivative! Once you compute it, you'll see that it combines both the derivative of \( 16x \) and the derivative of \( \cos(8x^2) \), which includes another application of the chain rule. Remember, it's always good to double-check your simplifications and calculations at each step to avoid small mistakes that can throw off your entire result! If you’re intrigued by trigonometric derivatives, consider exploring their relation to oscillatory motion in physics. You’ll find that functions like sine and cosine don’t just appear in calculus but also in waves, such as those created by sound or light. They are essential in modeling periodic phenomena and can lead to fascinating insights in both mathematics and science!

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