Solutions to Homework 2 SOSC 30100, Sept. 9, 2013

1. Here f(x) = 5x^{2} − 4 ∗ x the difference quotient in general is

\frac{f(x + Δx) − f(x)}{Δx} .

In this problem it is:

\frac{f(x + Δx) − f(x)}{Δx} = \frac{[5(x + Δx)^{2} − 4(x + Δx)] − [5x^{2} − 4x]}{Δx} .

Using the famous (a + b)^{2} = a^{2} + 2ab + b^{2} this equals

\frac{5x^{2} + 10(Δx)x + 5(Δx)^{2} − 4x − 4(Δx) − 5x^{2} + 4x}{Δx} .

Simplifying, this equals:

\frac{10(Δx)x + 5(Δx)^{2} − 4(Δx)}{Δx} .

Factoring the top, this equals:

\frac{(Δx)(10x + 5(Δx) − 4)}{Δx} .

Cancelling out Δx from top and bottom gives that this equals 10x − 4 + 5Δx .

Ans. to 1a: The difference quotient is 2x − 4 + 5Δx.

(b) The derivative is the limit as Δx → 0.

\frac{dy}{dx} = lim_{Δx→0} \frac{f(x + Δx) − f(x)}{Δx} . = lim_{Δx→0} 10x − 4 + 5Δx = 10x − 4 + 5 · 0 = 10x − 4 .

Ans. to 1b: The derivative dy/dx is 10x − 4. Sol. to 1c: Since f′(x) = 10x − 4, we have f′(2) = 10 · 2 − 4 = 20 − 4 = 16 and f′(3) = 10 · 3 − 4 = 30 − 4 = 26 and 2 Using the famous formula, (a + b)^{3} = a^{3} + 3a^{2}b + 3ab^{2} + b^{3} we can simplify q

q = \frac{(v + 2)^{3} − 8}{v} = \frac{v^{3} + 3v^{2}(2) + 3v(2^{2}) + 2^{3} − 8}{v} = \frac{v^{3} + 6v^{2} + 12v}{v} = \frac{(v^{2} + 6v + 12)v}{v} = v^{2}+6v+12 .

This is a nice polynomial in v so to find the limit at a point, just plug-it in a. lim_{v→0} q = 0^{2} + 6 · 0 + 12 = 12 b. lim_{v→2} q = 2^{2} + 6 · 2 + 12 = 4 + 12 + 12 = 28 c. lim_{v→a} q = a^{2} + 6 · a + 12 = 4 + 12 + 12 = 28

3. We use the constant multiple rule (you can always take a constant out of the differentiation) and the famous rule

\frac{d}{dx}x^{n} = nx^{n−1}

a.

\frac{d}{dx} (−x^{−4}) = − (\frac{d}{dx}x^{−4}) = −(−4)x^{−5} = 4x^{−5} .

b.

\frac{d}{du}(au^{b}) = a(\frac{d}{du}u^{b} = ab u^{b−1} .

c.

\frac{d}{dx}cx^{2} = c(\frac{d}{dx}x^{2} = c(2x) = 2c x .

d.

\frac{d}{dx}(7x^{1/3}) = 7(\frac{d}{dx}x^{1/3}) = 7 · (1/3)x^{−2/3} = \frac{7}{3}x^{− \frac{2}{3}} .

4. The product rule is (f(x)g(x))′ = f′(x)g(x) + f(x)g′(x). a. ((9x^{2} − 1)(3x + 1))′ = (9x^{2} − 1)′(3x + 1) + (9x^{2} − 1)(3x + 1)′ = (18x)(3x + 1) + (9x^{2} − 1)3 = 54x^{2} + 18x + 27x^{2} − 3 = 81x^{2} + 18x − 3 b. ((ax−b)(cx^{2}))′ = (ax−b)′(cx^{2})+(ax−b)(cx^{2})′ = (a)(cx^{2})+(ax−b)(2cx) = ac x^{2}+2ac x^{2}−2bc x = 3ac x^{2} − 2bc x c. [(2−3x)(1+x)(x+2)]′ = (2−3x)′(1+x)(x+2)+(2−3x)(1+x)′(x+2)+(2−3x)(1+x)(x+2)′ = (−3)(1+x)(x+2)+(2−3x)(1)(x+2)+(2−3x)(1+x)(1)

= −3(1 + x)(x + 2) + (2 − 3x)(x + 2) + (2 − 3x)(1 + x) .

d. [(x^{2}+3)x^{−1}]′ = (x^{2}+3)′(x^{−1})+(x^{2}+3)(x^{−1})′ = (2x)(x^{−1})+(x^{2}+3)(−1)x^{−2} = 2−(1+3x^{−2}) = 1−3x^{−2} .

5. For (a) and (b) it is better to first simplify, and then differentiate. For (c) and (d) we need the quotient rule. a.

((x^{2} + 3)/x)′ = (x + 3/x)′ = (x + 3x^{−1})′ = x′ + 3(x^{−1})′ = 1 + 3(−1)x^{−2} = 1 − \frac{3}{x^{2}} .

b.

((x + 7)/x)′ = (1 + 7/x)′ = (1 + 7x^{−1})′ = 1′ + 7(x^{−1})′ = 0 + 7(−1)x^{−2} = −\frac{7}{x^{2}} .

The quotient rule is:

(\frac{f(x)}{g(x)})′ = \frac{f′(x)g(x) − f(x)g′(x)}{g(x)^{2}} .

c.

(\frac{4x}{x + 5})′ = \frac{(4x)′(x + 5) − 4x(x + 5)′}{(x + 5)^{2}} = \frac{4(x + 5) − 4x(1)}{(x + 5)^{2}} = \frac{4x + 20 − 4x}{(x + 5)^{2}} = \frac{20}{(x + 5)^{2}} .

d.

(\frac{ax^{2} + b}{cx + d})′ = \frac{(ax^{2} + b)′(cx + d) − (ax^{2} + b)(cx + d)′}{(cx + d)^{2}} = \frac{(2ax)(cx + d) − (ax^{2} + b)(c)}{(cx + d)^{2}}

= \frac{2ac x^{2} + 2ad x − ac x^{2} − bc}{(cx + d)^{2}} = \frac{ac x^{2} + 2ad x − bc}{(cx + d)^{2}} .

6. The chain rule says

\frac{dw}{dx} = \frac{dw}{dy} · \frac{dy}{dx} .

Here

\frac{dw}{dy} = 2ay , \frac{dy}{dx} = 2bx ,

Hence

\frac{dw}{dx} = (2ay)(2bx) = 4abxy .

Finally we replace y by what it stands for in terms of x, getting

\frac{dw}{dx} = 4abxy = 2abx(bx^{2} + cx) = 2abx^{2}(bx + c) .

7. ((16x + 3)^{−2})′ = (−2)((16x + 3)^{−3}) · (16x + 3)′ = (−2)((16x + 3)^{−3}) · (16) = −32(16x + 3)^{−3} .

By the quotient rule:

(\frac{1}{(16x + 3)^{2}} )′ = \frac{1′(16x + 3)^{2} − 1((16x + 3)^{2})′}{(16x + 3)^{4}} = \frac{0 · (16x + 3)^{2} − 1((2)16x + 3))(16)}{(16x + 3)^{4}} = \frac{−32}{(16x + 3)^{3}} = −32(16x+3)^{−3} .

The answers are identical because God (or Newton) said so! a. When x is VERY BIG −7x + 5 and −1 are insignifcant compared to 2x^{2} and x respectively, hence it is possible to only retain the leading powers at the top and bottom.

lim_{x→∞} \frac{x − 1}{2x^{2} − 7x + 5} = lim_{x→∞} \frac{x}{2x^{2}} = lim_{x→∞} \frac{1}{2x} = \frac{1}{2∞} = 0 .

Ans. to 8a: 0. ¯. Since the top and bottom happen to be 0 at x = 1, we can use L’Hôpital’s rule:

lim_{x→1} \frac{x − 1}{2x^{2} − 7x + 5} = lim_{x→1} \frac{(x − 1)′}{(2x^{2} − 7x + 5)′} = lim_{x→1} \frac{1}{4x − 7} = \frac{1}{4 · 1 − 7} = −\frac{1}{3} .

Ans. to 8b: −\frac{1}{3}. Comment: We can also do it only using algebra

lim_{x→1} \frac{x − 1}{2x^{2} − 7x + 5} = lim_{x→1} \frac{x − 1}{(x − 1)(2x − 5)} = lim_{x→1} \frac{1}{2x − 5} = \frac{1}{2 · 1 − 5} = −\frac{1}{3} .

9. First way: Using algebra, f(x) = x + 1 when x ≠ 1. Since f(1) = 2 the function given is a coplicated way of writing down f(x) = x + 1 that is well-known to be continuous. Second Way: f(x) is continuous when x ≠ 1 (since it is a quotient of two continuous functions, and the denominator does not vanish). The limit of f(x) as x goes to 1 happens to be 2 (either by L’Hôpital or using algebra). Since the limit of f(x) as x goes to 1 equals f(1), it is also continuous at x = 1, hence it is continuous over all the real line.

10. By the product and chain rules:

f′(x) = (\frac{1}{2}(x^{2} + 4x − 9)^{3}x^{−2})′ = \frac{1}{2}((x^{2} + 4x − 9)^{3}x^{−2})′ =

\frac{1}{2}[((x^{2}+4x−9)^{3})′x^{−2}+((x^{2}+4x−9)^{3})(x^{−2})′] = \frac{1}{2}[(3(x^{2}+4x−9)^{2}(2x+4)x^{−2}+((x^{2}+4x−9)^{3})(−2)x^{−3}]quad.

I hope that you don not expect me to simplify!