극한·미분 심화 전체 문제

Let $\theta = \arctan x$. Then $\theta$ is the angle of a right triangle that gives $\tan \theta = x$. In particular, the ratio of the opposite side to the adjacent side is $x$. So, we have a triangle that looks like this:

where the length of the hypotenuse came from the Pythagorean Theorem. Now, $$\sin\left(\arctan x\right) = \sin \theta = \frac{\text{opp}}{\text{hyp}} = \frac{x}{\sqrt{x^2+1}}$$ From here, we differentiate using the quotient rule:

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x}\left\{\frac{x}{\sqrt{x^2+1}} \right\}&= \frac{\sqrt{x^2+1}-x\frac{2x}{2\sqrt{x^2+1}}}{x^2+1}\\ &=\left(\frac{\sqrt{x^2+1}-\frac{x^2}{\sqrt{x^2+1}}}{x^2+1}\right)\cdot\frac{\sqrt{x^2+1}}{\sqrt{x^2+1}}\\ &=\frac{(x^2+1)-x^2}{(x^2+1)^{3/2}}\\ &=\frac{1}{(x^2+1)^{3/2}}=(x^2+1)^{-3/2} \end{aligned}$$

Remark: another strategy is to differentiate first, using the chain rule, then draw a triangle to simplify the resulting expression $\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}\left\{\sin\left(\arctan x\right)\right\}=\dfrac{\cos(\arctan x)}{1+x^2}$.

Let $\theta = \arcsin x$. Then $\theta$ is the angle of a right triangle that gives $\sin \theta = x$. In particular, the ratio of the opposite side to the hypotenuse is $x$. So, we have a triangle that looks like this:

where the length of the adjacent side came from the Pythagorean Theorem. Now, $$\cot\left(\arcsin x\right) = \cot \theta = \frac{\text{adj}}{\text{opp}} = \frac{\sqrt{1-x^2}}{x}$$ From here, we differentiate using the quotient rule:

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x}\left\{\frac{\sqrt{1-x^2}}{x} \right\}&= \frac{x\frac{-2x}{2\sqrt{1-x^2}}-\sqrt{1-x^2}}{x^2}\\ &=\frac{-x^2-(1-x^2)}{x^2\sqrt{1-x^2}}\\ &=\frac{-1}{x^2\sqrt{1-x^2}} \end{aligned}$$

Remark: another strategy is to differentiate first, using the chain rule, then draw a triangle to simplify the resulting expression $\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}\left\{\cot\left(\arcsin x\right)\right\}=\frac{-\csc^2(\arcsin x)}{\sqrt{1-x^2}}$.