Second Derivative Tan Inverse 08Cc52

1. **State the problem:** Find the second derivative of the function $$y = \tan^{-1}(\sqrt{x})$$. 2. **Recall the formula:** The derivative of $$\tan^{-1}(u)$$ with respect to $$x$$ is $$\frac{d}{dx} \tan^{-1}(u) = \frac{1}{1+u^2} \cdot \frac{du}{dx}$$. 3. **First, find the first derivative:** Let $$u = \sqrt{x} = x^{1/2}$$. Then $$\frac{du}{dx} = \frac{1}{2} x^{-1/2} = \frac{1}{2\sqrt{x}}$$. So, $$\frac{dy}{dx} = \frac{1}{1+u^2} \cdot \frac{du}{dx} = \frac{1}{1+x} \cdot \frac{1}{2\sqrt{x}} = \frac{1}{2\sqrt{x}(1+x)}$$. 4. **Find the second derivative:** We differentiate $$\frac{dy}{dx} = \frac{1}{2\sqrt{x}(1+x)}$$. Rewrite as: $$\frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{x^{1/2}(1+x)} = \frac{1}{2} x^{-1/2} (1+x)^{-1}$$. Use the product rule: $$\frac{d^2y}{dx^2} = \frac{1}{2} \left[ \frac{d}{dx} (x^{-1/2}) \cdot (1+x)^{-1} + x^{-1/2} \cdot \frac{d}{dx} (1+x)^{-1} \right]$$. Calculate derivatives: $$\frac{d}{dx} (x^{-1/2}) = -\frac{1}{2} x^{-3/2}$$, $$\frac{d}{dx} (1+x)^{-1} = - (1+x)^{-2}$$. Substitute: $$\frac{d^2y}{dx^2} = \frac{1}{2} \left[ -\frac{1}{2} x^{-3/2} (1+x)^{-1} + x^{-1/2} (- (1+x)^{-2}) \right]$$ Simplify: $$= \frac{1}{2} \left[ -\frac{1}{2} x^{-3/2} (1+x)^{-1} - x^{-1/2} (1+x)^{-2} \right] = -\frac{1}{4} x^{-3/2} (1+x)^{-1} - \frac{1}{2} x^{-1/2} (1+x)^{-2}$$. 5. **Final answer:** $$\boxed{\frac{d^2y}{dx^2} = -\frac{1}{4 x^{3/2} (1+x)} - \frac{1}{2 x^{1/2} (1+x)^2}}$$. This is the second derivative of $$y = \tan^{-1}(\sqrt{x})$$.