극한·미분 심화 전체 문제

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 $\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}\left\{\cot\left(\arcsin x\right)\right\}=\frac{-\csc^2(\arcsin x)}{\sqrt{1-x^2}}$.

(a) Since $e$ raised to any power is always positive (and thus never $0$), it follows that the only value that makes $f'(x) = (x-1)e^{-x^2} = 0$ is $x=1$. This is confirmed by plotting $f'$ where we only see the graph cross the $x$-axis at $x=1$ (though we do see the graph approach the $x$-axis) as $x$ increases or decreases without bound). (b) We know that $f''(x) > 0$ wherever $f'$ is increasing. From the graph of $f'$, we see that $f'$ is increasing approximately on the interval $-0.37 < x < 1.37$, and thus $f$ is concave up on this interval. (c) The tangent line approximation at $a = 2$ is $L(x) = f(2) + f'(2)(x-2) \approx -3 + 0.01832(x-2)$. Thus, $f(1.88) \approx L(1.88) \approx -3 + 0.01832(1.88-2) = -3.00022$. Since $f''(x) < 0$ at and around $x = 2$, $f$ is concave down there, and therefore the tangent line lies above the curve. This makes our estimate larger than the true value of $f(1.88)$.