Derivative of $\arctan\big (s-\sqrt{1+s^{2}}\big) $

Question

I am supposed to find the derivative of $f(s) = \arctan\big(s-\sqrt{1+s^{2}}\big) $. My first step was this: $\frac{1}{1+ (s-\sqrt{1+s^{2}})^{2} } $. What am I supposed to do next? Thanks

Answer 1

Use the Chain Rule.

$$\frac{d}{ds} \arctan u = \frac{1}{1+u^2}\cdot \color{blue}{\frac{du}{ds}} \tag{1}$$

Let $u = s-\sqrt{1+s^2}$ and solve for $\color{blue}{\frac{du}{ds}}$.

$$\color{blue}{\frac{du}{ds} = \frac{d}{ds}\big(s-\sqrt{1+s^2}\big)}$$

$$\color{blue}{= 1-\frac{d}{ds}\sqrt{1+s^2}}$$

Use the Chain Rule once more to find $\color{purple}{\frac{d}{ds}\sqrt{1+s^2}}$.

$$\color{purple}{\frac{d}{ds}\sqrt{1+s^2} = \frac{1}{2\sqrt{1+s^2}}\cdot2s}$$

Returning to the blue expression, you get

$$\color{blue}{= 1-\frac{2s}{2\sqrt{1+s^2}} = 1-\frac{s}{\sqrt{1+s^2}}} \tag{2}$$

Returning to $(1)$, you get

$$\frac{d}{ds}\arctan\big(s-\sqrt{1+s^2}\big) = \frac{1}{1+\big(s-\sqrt{1+s^2}\big)^2}\cdot\bigg(1-\frac{s}{\sqrt{1+s^2}}\bigg)$$

Which can be further simplified (not very easily and neatly).

$$\frac{1}{1+\big(s-\sqrt{1+s^2}\big)^2}\cdot\bigg(1-\frac{s}{\sqrt{1+s^2}}\bigg) = \frac{1}{1+\big(s-\sqrt{1+s^2}\big)^2}\cdot\bigg(\frac{\sqrt{1+s^2}-s}{\sqrt{1+s^2}}\bigg)$$

$$= \frac{(\sqrt{1+s^2}-s}{\big(1+\big(s-\sqrt{1+s^2}\big)^2\big)\cdot\big(\sqrt{1+s^2}\big)}$$

$$= \frac{\sqrt{1+s^2}-s}{\big(1+s^2-2s\sqrt{1+s^2}+1+s^2\big)\cdot\big(\sqrt{1+s^2}\big)}$$

$$= \frac{\sqrt{1+s^2}-s}{\big(2+2s^2-2s\sqrt{1+s^2}\big)\cdot\big(\sqrt{1+s^2}\big)}$$

$$= \frac{\sqrt{1+s^2}-s}{2\sqrt{1+s^2}+2s^2\sqrt{1+s^2}-2s(1+s^2)}$$

$$= \frac{\sqrt{1+s^2}-s}{2\big(\sqrt{1+s^2}+s^2\sqrt{1+s^2}-s(1+s^2)\big)}$$

$$= \frac{\sqrt{1+s^2}-s}{2\big(\sqrt{1+s^2}(1+s^2)-s(1+s^2)\big)}$$

$$= \frac{\sqrt{1+s^2}-s}{2\big((1+s^2)\big(\sqrt{1+s^2}-s\big)\big)} = \frac{1}{2(s^2+1)} = \frac{1}{2s^2+2}$$

Answer 2

May be this method is easier:

$$y=arctan(s-\sqrt{1+s^2})$$

$$\tan y=(s-\sqrt{1+s^2})$$

Differentiating both sides we get:

$$(1+\tan^2 y) dy=ds-\frac{s}{\sqrt{1+s^2}} ds$$

$$y'=\frac{dy}{ds}=\frac{1}{1+(s-\sqrt{1+s^2})^2}.(1-\frac{s}{\sqrt{1+s^2}})$$