Question: Find the derivative of the function.
y = 3 \tan^{-1}(x - \sqrt{1 + x^2})
y' =
1. **State the problem:** Find the derivative of the function $$y = 3 \tan^{-1}\left(x - \sqrt{1 + x^2}\right)$$.
2. **Recall the derivative formula:** The derivative of $$\tan^{-1}(u)$$ with respect to $$x$$ is $$\frac{du/dx}{1 + u^2}$$.
3. **Define the inner function:** Let $$u = x - \sqrt{1 + x^2}$$.
4. **Find $$du/dx$$:**
$$\frac{du}{dx} = 1 - \frac{d}{dx}\left(\sqrt{1 + x^2}\right)$$
The derivative of $$\sqrt{1 + x^2}$$ is:
$$\frac{d}{dx}\left(\sqrt{1 + x^2}\right) = \frac{1}{2\sqrt{1 + x^2}} \cdot 2x = \frac{x}{\sqrt{1 + x^2}}$$
So,
$$\frac{du}{dx} = 1 - \frac{x}{\sqrt{1 + x^2}}$$
5. **Simplify $$du/dx$$:**
$$\frac{du}{dx} = \frac{\sqrt{1 + x^2} - x}{\sqrt{1 + x^2}}$$
6. **Calculate $$1 + u^2$$:**
$$u = x - \sqrt{1 + x^2}$$
Square it:
$$u^2 = \left(x - \sqrt{1 + x^2}\right)^2 = x^2 - 2x\sqrt{1 + x^2} + (1 + x^2) = 2x^2 + 1 - 2x\sqrt{1 + x^2}$$
Then,
$$1 + u^2 = 1 + 2x^2 + 1 - 2x\sqrt{1 + x^2} = 2 + 2x^2 - 2x\sqrt{1 + x^2}$$
Factor out 2:
$$1 + u^2 = 2\left(1 + x^2 - x\sqrt{1 + x^2}\right)$$
7. **Write the derivative $$y'$$:**
$$y' = 3 \cdot \frac{du/dx}{1 + u^2} = 3 \cdot \frac{\frac{\sqrt{1 + x^2} - x}{\sqrt{1 + x^2}}}{2\left(1 + x^2 - x\sqrt{1 + x^2}\right)}$$
8. **Simplify the fraction:**
$$y' = 3 \cdot \frac{\sqrt{1 + x^2} - x}{\sqrt{1 + x^2}} \cdot \frac{1}{2\left(1 + x^2 - x\sqrt{1 + x^2}\right)} = \frac{3}{2} \cdot \frac{\sqrt{1 + x^2} - x}{\sqrt{1 + x^2} \left(1 + x^2 - x\sqrt{1 + x^2}\right)}$$
9. **Final answer:**
$$\boxed{y' = \frac{3}{2} \cdot \frac{\sqrt{1 + x^2} - x}{\sqrt{1 + x^2} \left(1 + x^2 - x\sqrt{1 + x^2}\right)}}$$