Calculus I, Fall 2021

Brief solutions to Midterm 01

Nov 02, 2021

1. (6 pts) Find lim_{y→+∞}y sin \frac{2}{\sqrt{y}}.

Ans:

lim_{y→+∞} y sin \frac{2}{\sqrt{y}} = lim_{y→+∞} \sqrt{y} \frac{sin \frac{2}{\sqrt{y}}}{\frac{1}{\sqrt{y}}} = lim_{y→+∞} \sqrt{y} · lim_{x→0} \frac{sin(2x)}{x} = ∞ · 2 = ∞

2. (10 pts) Evaluate lim_{z→x} \frac{sec(z^{2} + 1) − sec(x^{2} + 1)}{e^{z} − e^{x}}. Hint: what is the definition of f′(x)?

Ans:

lim_{z→x} \frac{sec(z^{2} + 1) − sec(x^{2} + 1)}{e^{z} − e^{x}} = lim_{z→x} \frac{sec(z^{2} + 1) − sec(x^{2} + 1)}{z − x} lim_{z→x} \frac{z − x}{e^{z} − e^{x}}

= lim_{z→x} \frac{sec(z^{2} + 1) − sec(x^{2} + 1)}{z − x} ; lim_{z→x} \frac{e^{z} − e^{x}}{z − x} = \frac{\frac{d}{dx} sec(x^{2} + 1)}{\frac{d}{dx}e^{x}} (5pts)

= \frac{sec(x^{2} + 1) · tan(x^{2} + 1) · 2x}{e^{x}} (5pts)

3. (4+4+8 pts)

(a) Write down precise definition of lim_{x→c} f(x) = L

(b) State the Sandwich Theorem.

(c) Prove the Sandwich Theorem using the definition in (a). Ans: (a),(b): see textbook. (c): Suppose g(x) ≤ f(x) ≤ h(x) on (c−a.c)∪(c, c+a) (that is, {x | 0 < |x−c| < a}), a > 0 and lim_{x→c} g(x) = lim_{x→c} h(x) = L. Then for any ε > 0, there exists a δ_{1} > 0 and a δ_{2} > 0 such that

0 < |x − c| < δ_{1} ⟹ |g(x) − L| < ε ⟹ L − ε < g(x)

and

0 < |x − c| < δ_{2} ⟹ |h(x) − L| < ε ⟹ h(x) < L + ε

Take δ = min{δ_{1}, δ_{2}, a}. It follows that 0 < |x − c| < δ ⟹ L − ε < g(x) ≤ f(x) ≤ h(x) < L + ε ⟹ |f(x) − L| < ε.

4. (2+8 pts) True or False? Prove it if true, find a counter example if false. If f(x) is differentiable at x = c, then it is continuous at x = c. Ans: True. See textbook for the proof.

5. (12 pts) Let f_{i}(x) = a_{i}x^{2} + b_{i}x + c_{i}, i = 1, 2, 3. Evaluate \frac{d}{dx} | \begin{matrix}f_{1}(x) & f′_{1}(x) & f′′_{1}(x) \\ f_{2}(x) & f′_{2}(x) & f′′_{2}(x) \\ f_{3}(x) & f′_{3}(x) & f′′_{3}(x) \end{matrix} | .

Ans:

= | \begin{matrix}f′_{1}(x) & f′_{1}(x) & f′′_{1}(x) \\ f′_{2}(x) & f′_{2}(x) & f′′_{2}(x) \\ f′_{3}(x) & f′_{3}(x) & f′′_{3}(x) \end{matrix} | + | \begin{matrix}f_{1}(x) & f′′_{1}(x) & f′′_{1}(x) \\ f_{2}(x) & f′′_{2}(x) & f′′_{2}(x) \\ f_{3}(x) & f′′_{3}(x) & f′′_{3}(x) \end{matrix} | + | \begin{matrix}f_{1}(x) & f′_{1}(x) & f′′′_{1}(x) \\ f_{2}(x) & f′_{2}(x) & f′′′_{2}(x) \\ f_{3}(x) & f′_{3}(x) & f′′′_{3}(x) \end{matrix} | = 0

The first 2 determinants are zero since they both have identical columns. The third determinant is zero since f′′′_{i}(x) = 0.

6. (12 pts) Find the smallest positive integer n such that \frac{d^{n}}{dx^{n}} (x^{10} sin x)|_{x=0} is nonzero and find this value. Explain. Ans: Since

\frac{d^{n}}{dx^{n}} (x^{10} sin x)|_{x=0} = ∑_{k=0}^{n} \binom{n}{k}\frac{d^{k}}{dx^{k}} x^{10}|_{x=0} \frac{d^{n−k}}{dx^{n−k}} sin x|_{x=0}

and

\frac{d^{p}}{dx^{p}} x^{10}|_{x=0} ≠ 0 ⟺ p = 10, \frac{d^{q}}{dx^{q}} sin x|_{x=0} ≠ 0 ⟺ q = 1, 3, 5, ⋯,

It follows that the smallest n = 11 (p = 10, q = 1). Answer = \binom{11}{10} · 10! · cos 0 = 11!

7. (10 pts) Evaluate \frac{d}{dx}(tan x)^{\sqrt{x}}, 0 < x < \frac{π}{2}.

Ans:

(tan x)^{\sqrt{x}} = (e^{ln(tan x)})^{\sqrt{x}} = e^{\sqrt{x} ln(tan x)},

therefore, the answer is (need not simplify)

\frac{d}{dx}e^{\sqrt{x} ln(tan x)} = (tan x)^{\sqrt{x}} \frac{d}{dx}(\sqrt{x} ln(tan x)) = (tan x)^{\sqrt{x}} (\frac{1}{2\sqrt{x}} ln(tan x) + \sqrt{x}\frac{sec^{2} x}{tan x})

8. (12 pts) Find the normal line and tangent line of x^{4} + y^{2} = 2 at (1, −1).

Ans:

x^{4} + y^{2} = 2 ⟹ 4x^{3} + 2yy′ = 0 ⟹ y′ = −2\frac{x^{3}}{y} = 2 (4pts)

tangent line : \frac{y + 1}{x − 1} = 2, (4pts) normal line : \frac{y + 1}{x − 1} = −\frac{1}{2}. (4pts)

9. (4+8 pts) Let f^{−1} be the inverse function of f. Suppose f and f^{−1} are twice differentiable (i.e. both first and second derivative exist).

(a) \underline{Derive} the formula of \frac{d}{dy} f^{−1}(y) as a function of y in terms of f′(·) and f^{−1}(·).

(b) \underline{Derive} the formula of \frac{d^{2}}{dy^{2}} f^{−1}(y) as a function of y in terms of f′(·), f′′(·) and f^{−1}(·).

Ans: Denote by y = f(x), or x = f^{−1}(y). (a):

f^{−1}(f(x)) = x ⟹ (f^{−1})′(y) · f′(x) = 1 ⟹ (f^{−1})′(y) = \frac{1}{f′(x)} = \frac{1}{f′(f^{−1}(y))}

(b):

\frac{d}{dx}((f^{−1})′(y) · f′(x) = 1) ⟹ (f^{−1})′′(y) · (f′(x))^{2} + (f^{−1})′(y) · f′′(x) = 0

⟹ (f^{−1})′′(y) = \frac{−(f^{−1})′(y) · f′′(x)}{(f′(x))^{2}} = \frac{−f′′(x)}{(f′(x))^{3}} = \frac{−f′′(f^{−1}(y))}{(f′(f^{−1}(y)))^{3}}

Method 2 for (b): take \frac{d}{dx} on (a) and use quotient rule and the Chain Rule.