# What is the second derivative of #f(x)=sin(1/x^2) #?

Find the first derivative using chain rule. To find the second derivative use the product rule. So separate the functions into f and g then find f' and g' and then apply the product rule.

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To find the second derivative of ( f(x) = \sin\left(\frac{1}{x^2}\right) ), you need to first find the first derivative and then differentiate it again.

- Find the first derivative: [ f'(x) = \frac{d}{dx}\left[\sin\left(\frac{1}{x^2}\right)\right] ]

Using the chain rule, the derivative of (\sin(u)) is (\cos(u)) multiplied by the derivative of (u): [ f'(x) = \cos\left(\frac{1}{x^2}\right) \cdot \frac{d}{dx}\left(\frac{1}{x^2}\right) ]

Differentiate (\frac{1}{x^2}): [ \frac{d}{dx}\left(\frac{1}{x^2}\right) = -2x^{-3} ]

[ f'(x) = \cos\left(\frac{1}{x^2}\right) \cdot (-2x^{-3}) ] [ f'(x) = -2x^{-3} \cos\left(\frac{1}{x^2}\right) ]

- Now, find the second derivative by differentiating (f'(x)) with respect to (x): [ f''(x) = \frac{d}{dx}\left[-2x^{-3} \cos\left(\frac{1}{x^2}\right)\right] ]

Using the product rule and the chain rule: [ f''(x) = -2 \cdot (-3)x^{-4} \cos\left(\frac{1}{x^2}\right) + (-2x^{-3}) \cdot \frac{d}{dx}\left(\cos\left(\frac{1}{x^2}\right)\right) ]

[ f''(x) = 6x^{-4} \cos\left(\frac{1}{x^2}\right) + 2x^{-3} \sin\left(\frac{1}{x^2}\right) \cdot \frac{d}{dx}\left(\frac{1}{x^2}\right) ]

Differentiate (\frac{1}{x^2}) to get: [ \frac{d}{dx}\left(\frac{1}{x^2}\right) = -2x^{-3} ]

[ f''(x) = 6x^{-4} \cos\left(\frac{1}{x^2}\right) + 2x^{-3} \sin\left(\frac{1}{x^2}\right) \cdot (-2x^{-3}) ]

[ f''(x) = 6x^{-4} \cos\left(\frac{1}{x^2}\right) - 4x^{-6} \sin\left(\frac{1}{x^2}\right) ]

Therefore, the second derivative of ( f(x) = \sin\left(\frac{1}{x^2}\right) ) is ( f''(x) = 6x^{-4} \cos\left(\frac{1}{x^2}\right) - 4x^{-6} \sin\left(\frac{1}{x^2}\right) ).

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