1. **State the problem:** We need to find the derivative of the function $$h(x) = \arctan(\sin(\frac{1}{x^2}))$$. 2. **Recall the chain rule and derivative formulas:** - The derivative of $$\arctan(u)$$ with respect to $$x$$ is $$\frac{1}{1+u^2} \cdot \frac{du}{dx}$$. - The derivative of $$\sin(v)$$ with respect to $$x$$ is $$\cos(v) \cdot \frac{dv}{dx}$$. 3. **Identify inner functions:** - Let $$u = \sin(\frac{1}{x^2})$$. - Let $$v = \frac{1}{x^2} = x^{-2}$$. 4. **Compute derivatives step-by-step:** - $$\frac{dv}{dx} = \frac{d}{dx} x^{-2} = -2x^{-3} = -\frac{2}{x^3}$$. - $$\frac{du}{dx} = \cos(v) \cdot \frac{dv}{dx} = \cos\left(\frac{1}{x^2}\right) \cdot \left(-\frac{2}{x^3}\right) = -\frac{2}{x^3} \cos\left(\frac{1}{x^2}\right)$$. 5. **Apply the chain rule for $$h(x)$$:** $$ \frac{dh}{dx} = \frac{1}{1 + u^2} \cdot \frac{du}{dx} = \frac{1}{1 + \sin^2\left(\frac{1}{x^2}\right)} \cdot \left(-\frac{2}{x^3} \cos\left(\frac{1}{x^2}\right)\right) $$ 6. **Final derivative expression:** $$ \boxed{h'(x) = -\frac{2 \cos\left(\frac{1}{x^2}\right)}{x^3 \left(1 + \sin^2\left(\frac{1}{x^2}\right)\right)}} $$ This derivative shows how the rate of change of $$h(x)$$ depends on $$x$$, with oscillations becoming more rapid near zero due to the $$\frac{1}{x^2}$$ term inside the sine and cosine functions.