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tags: Calculus
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# Calculus B. Ⅱ Midterm 110 Notes

## Power formula

### State

If $y=f(x)=x^r,x>0,r\in\mathbb{R}$, then ${dy\over dx}=f'(x)=rx^{r-1}$.

### Pf.

By _Theorem 6.4.3_ and _Proposition 6.5.4_,

$$
{dy\over dx}={dx^r\over dx}={de^{r\ln x}\over dx}=e^{r\ln x}{r\over x}=rx^{r-1}
$$

## $\csc^{-1}$

If we restrict $x$ on $[{-\pi\over2},{\pi\over2}]\setminus\{0\}$, then $y=f(x)=\csc x$ is bijective, i.e. $f:[{-\pi\over2},{\pi\over2}]\setminus\{0\}\mapsto\mathbb{R}\setminus(-1, 1)$ is bijective. Hence by _Preposition 6.1.2_, we have the inverse function $x=f^{-1}(y)=\csc^{-1}(y)$, where $f^{-1}:\mathbb{R}\setminus(-1, 1)\mapsto[{-\pi\over2},{\pi\over2}]\setminus\{0\}$. So we have
$$g:\mathbb{R}\setminus(-1, 1)\mapsto[{-\pi\over2},{\pi\over2}]\setminus\{0\},y=g(x)=\csc^{-1}x$$.

Then since $y=\csc^{-1}x,\csc y=x$, we have $-\csc y\cot y{dy\over dx}=1$. Moreover, since $\cot^2y=\csc^2y-1=x^2-1$, we have the two cases:
1. If $y\in(0,{\pi\over2}]$, then $x\geq1$ and $\cot y=\sqrt{x^2-1}$. So we have $${dy\over dx}={d\csc^{-1}x\over dx}={-1\over x\sqrt{x^2-1}}, x\ne1$$
2. If $y\in[{-\pi\over2}, 0)$, then $x\leq-1$ and $\cot y=-\sqrt{x^2-1}$. So we have $${dy\over dx}={d\csc^{-1}x\over dx}={1\over x\sqrt{x^2-1}}, x\ne-1$$

Hence we have $${dy\over dx}={d\csc^{-1}x\over dx}={-1\over |x|\sqrt{x^2-1}}, |x|\ne1$$.

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