곱과 몫의 미분법

(a) Using the Product Rule and the information given, $$\begin{aligned}w'(t) &= \frac{d}{dt}\left(t^t\right)\cdot\left(\arccos t\right)+t^t\cdot\frac{d}{dt}\left(\arccos t\right) \\ &= t^t\left(\ln t+1\right)\cdot\left(\arccos t\right)+t^t\cdot\frac{-1}{\sqrt{1-t^2}}\end{aligned}$$ (b) Using the local linearization formula, we have $L(t) = w\left(\frac{1}{2}\right) + w'\left(\frac{1}{2}\right) \left(t-\frac{1}{2}\right)$. Substituting $t=\frac{1}{2}$ into the formulas for $w$ and $w'$ and using a computational device gives $$w(2) = \left(\frac{1}{2}\right)^{\left(\frac{1}{2}\right)}\arccos\left(\frac{1}{2}\right) \approx 0.740$$ and $$w'(2) = \left[\left(\frac{1}{2}\right)^{\left(\frac{1}{2}\right)}\left(\ln t+1\right)\cdot\left(\arccos \left(\frac{1}{2}\right)\right)+\left(\frac{1}{2}\right)^{\left(\frac{1}{2}\right)}\cdot\frac{-1}{\sqrt{1-\left(\frac{1}{2}\right)^2}}\right] \approx -0.589\,.$$ Thus, $L(t) \approx 0.740-0.589(t-0.5)$. (c) Using the Quotient Rule, $$v'(t) = frac{(t)^{t}\left(\ln(t)+1\right)\left(\arccos(t)\right)-(t)^{t}\cdot\left(\frac{-1}{\sqrt{1-(t)^2}}\right)}{\arccos^2(t)}\,.$$ Using a computational device to evaluate $v'(2)$, it follows $v'(2) \approx 0.952$. Since this number is positive, the instantaneous rate of change of $v$ at $t=\frac{1}{2}$ is positive, which means that $v$ is increasing at that instant.