Calculus I, Fall 2015 (http://www.math.nthu.edu.tw/˜ wangwc/)

Brief answer to selected problems in HW00

1. Section 3.8: Problem 98. Let

f(x) = ln x ⟹ f′(x) = \frac{1}{x}

By definition, for fixed x > 0

f′(1) = lim_{\frac{x}{n} →0} \frac{f(1+ \frac{x}{n})−f(1)}{\frac{x}{n}} = lim_{n→∞} \frac{n}{x}(ln(1 + \frac{x}{n}) − ln 1)
= \frac{1}{x} lim_{n→∞} ln(1 + \frac{x}{n})^{n}

which implies

x = ln( lim_{n→∞}(1 + \frac{x}{n})^{n}) ⟹ lim_{n→∞}(1 + \frac{x}{n})^{n} = e^{x}

2. Section 3.9: Problem 55. After differentiation, we find out

f′(x) = g′(x), x > 0

which means

f(x) = g(x) + c, x ≥ 0

where c is some constant. And x = 0 gives us

−\frac{π}{2} = 0 + c, x ≥ 0

hence

f(x) = g(x) − \frac{π}{2} , x ≥ 0

On the other hand, we also can derive same result by following process. For x ≥ 0, let

θ = sin^{−1} \frac{x − 1}{x + 1} ⟹ sin θ = \frac{x − 1}{x + 1}

then

tan(\frac{θ + \frac{π}{2}}{2} ) = \sqrt{\frac{1 − cos(θ + \frac{π}{2})}{1 + cos(θ + \frac{π}{2})}} = \sqrt{\frac{1 + sin θ}{1 − sin θ}} = \sqrt{x}

hence

tan^{−1} \sqrt{x} = \frac{θ + \frac{π}{2}}{2} ⟹ sin^{−1} \frac{x − 1}{x + 1} = 2 tan^{−1} \sqrt{x} − \frac{π}{2}

3. Hw05: Problem 3. Domain and range for csc and csc^{−1}

csc  : {θ|θ ∈ R, θ ≠ kπ, ∀k ∈ Z} ↦ (−∞, −1] ∪ [1, ∞)
restricted csc : [−\frac{π}{2} , 0) ∪ (0, \frac{π}{2} ] ↦ (−∞, −1] ∪ [1, ∞)
csc^{−1}  : (−∞, −1] ∪ [1, ∞) ↦ [−\frac{π}{2} , 0) ∪ (0, \frac{π}{2} ]

For |x| > 1, let

θ = csc^{−1} x ⟹ csc θ = x
⟹ − csc θ cot θ\frac{dθ}{dx} = 1
⟹ \frac{dθ}{dx} = \frac{−1}{csc θ cot θ}

since

cot θ \{ \begin{matrix}> 0 & , θ ∈ (0, \frac{π}{2}) or x > 1 \\ < 0 & , θ ∈ (−\frac{π}{2}, 0) or x < −1 \end{matrix}

we have

cot θ = \{ \begin{matrix}\sqrt{x^{2} − 1} & , x > 1 \\ −\sqrt{x^{2} − 1} & , x < −1 \end{matrix}

hence

\frac{d csc^{−1} x}{dx} = \frac{−1}{|x|\sqrt{x^{2} − 1}}, |x| > 1