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B.) Compute $\dfrac{\mathrm{d}}{\mathrm{d}x} \mathrm{arccsc(x)}$, where $\mathrm{arccsc}$ denotes the inverse cosecant function.
Solution: Let $\theta=\mathrm{arccsc}(x)$ so $\csc(\theta)=x$. Since $\dfrac{\mathrm{d}}{\mathrm{d}x}\csc(x)=-\cot(x)\csc(x)$, we see $$-\cot(\theta)\csc(\theta)\dfrac{\mathrm{d}\theta}{\mathrm{d}x}=1.$$ Therefore we get $$\dfrac{\mathrm{d}\theta}{\mathrm{d}x} = -\dfrac{1}{\cot(\theta)\csc(\theta)}.$$ First off, $\csc(\theta)=x$ so we get $$\dfrac{\mathrm{d}\theta}{\mathrm{d}x} = -\dfrac{1}{x\cot(\theta)}.$$ Now, consider $\csc(\theta)=x$ and remember that $\csc(\theta) =\dfrac{1}{\sin(\theta)}$. We draw the following triangle representing the situation:

We need the midding side in order to compute $\cot(\theta) = \dfrac{\mathrm{adjacent}}{\mathrm{hypotenuse}}$. To solve for the missing side, we solve for $?$ in the equation $$?^2 + 1^2 = x^2,$$ or $$? = \sqrt{x^2-1},$$ where we threw away the negative solution because it is not meaningful. Therefore $$\cot(\theta) = \dfrac{\sqrt{x^2-1}}{1},$$ and so we get $$\dfrac{\mathrm{d}\theta}{\mathrm{d}x} = -\dfrac{1}{x\cot(\theta)} = -\dfrac{1}{x\sqrt{x^2-1}}.$$

Section 2.6 #36: When a certain polyatomic gas undergoes adiabatic expansion, its pressure $p$ and volume $V$ satisfy the equation $$pV^{1.3}=k,$$ where $k$ is a constant. Find the relationship between the related rates $\dfrac{\mathrm{d}p}{\mathrm{d}t}$ and $\dfrac{\mathrm{d}V}{\mathrm{d}t}$.
Solution: Differentiate the equation to get $$\dfrac{\mathrm{d}p}{\mathrm{d}t} V^{1.3} + 1.3 p V^{0.3} \dfrac{\mathrm{d}V}{\mathrm{d}t}=0.$$ Therefore we see $$\dfrac{\mathrm{d}p}{\mathrm{d}t} = -1.3p \dfrac{V^{0.3}}{V^{1.3}} \dfrac{\mathrm{d}V}{\mathrm{d}t}=-1.3p V^{-0.3} \dfrac{\mathrm{d}V}{\mathrm{d}t}.$$ There are other acceptable expressions for "the relationship" in this problem, depending on which variables you wish to appear on the right-hand side.