다항함수의 미분법

(a) We note that $w$ has the structure $w(t) = 3r(t) - 2s(t)$, so by the sum and constant multiple rules, $w'(t) = 3r'(t) - 2s'(t)$. Applying the given information about $r'$ and $s'$, $w'(t) = 3t^t(\ln(t) + 1) + 2\frac{1}{\sqrt{1-t^2}}$. (b) We see that $w(\frac{1}{2}) = 3(\frac{1}{2})^{\frac{1}{2}} - 2\arccos(\frac{1}{2}) = 3 \cdot \frac{1}{\sqrt{2}} - 2 \cdot \frac{\pi}{3} = \frac{3}{\sqrt{2}} - \frac{2\pi}{3}$. Similarly, $$\begin{aligned}w'\left(\frac{1}{2}\right) &= 3\left(\frac{1}{2}\right)^{\frac{1}{2}}\left(\ln\left(\frac{1}{2}\right) + 1\right) + 2\frac{1}{\sqrt{1-(\frac{1}{2})^2}} \\ &= \frac{3}{\sqrt{2}}\left(\ln\left(\frac{1}{2}\right) + 1\right) + \frac{4}{\sqrt{3}}\end{aligned}\,.$$ Using these values, we find that the tangent line to $w$ at the point $(\frac{1}{2},w(\frac{1}{2})$ is $$L(t) = w\left(\frac{1}{2}\right) + w'\left(\frac{1}{2}\right)\left(t-\frac{1}{2}\right) = \left(\frac{3}{\sqrt{2}} - \frac{2\pi}{3}\right) + \left(\frac{3}{\sqrt{2}}\left(\ln\left(\frac{1}{2}\right) + 1\right) + \frac{4}{\sqrt{3}}\right)\left(t-\frac{1}{2}\right)\,.$$ (c) To determine whether $v$ is increasing or decreasing at $t = \frac{1}{2}$, we compute $v'(\frac{1}{2})$. First, $v'(t) = t^t(\ln(t) + 1) - \frac{1}{\sqrt{1-t^2}}$. Hence, $$v'\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^{\frac{1}{2}}\left(\ln\left(\frac{1}{2}\right) + 1\right) - \frac{1}{\sqrt{1-(\frac{1}{2})^2}} \approx -0.93772\,.$$ Since $v'(\frac{1}{2})$ is negative, $v$ is decreasing at $t = \frac{1}{2}$.