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\centerline
{
\bf
Solutions to Homework 2 SOSC 30100, Sept. 9, 2013
}


{\bf 1.} Here $f(x)=5x^2-4*x$ the {\bf difference quotient} in general is
$$
\frac{f(x+ \Delta x)-f(x)}{\Delta x} \quad .
$$
In this problem it is:
$$
\frac{f(x+ \Delta x)-f(x)}{\Delta x} =
\frac{[5(x+ \Delta x)^2- 4(x+\Delta x)]-  [5x^2-4x] }{\Delta x} \quad .
$$
Using the famous $(a+b)^2=a^2+2ab+b^2$ this equals
$$
\frac{5x^2+ 10 (\Delta x)x + 5 (\Delta x)^2- 4x - 4 (\Delta x) -  5x^2 + 4x }{\Delta x} \quad .
$$
Simplifying, this equals:
$$
\frac{10 (\Delta x)x + 5 (\Delta x)^2 - 4 (\Delta x) }{\Delta x} \quad .
$$
Factoring the top, this equals:
$$
\frac{(\Delta x) (10x + 5(\Delta x) - 4) }{\Delta x} \quad .
$$
Cancelling out $\Delta x$ from top and bottom gives that this equals
$$
10x -4 +5 \Delta x   \quad .
$$

{\bf Ans. to 1a}: The difference quotient is $2x-4 + 5 \Delta x$.

(b) The derivative is the {\bf limit} as $\Delta x \rightarrow 0$. 
$$
\frac{dy}{dx} = \lim_{\Delta x \rightarrow 0} \,\, \frac{f(x+ \Delta x)-f(x)}{\Delta x} \quad .
= \lim_{\Delta x \rightarrow 0} 10x -4 +5 \Delta x = 10x -4 +5 \cdot 0 = 10x-4 \quad .
$$

{\bf Ans. to 1b}: The derivative $dy/dx$ is $10x-4$.

{\bf Sol. to 1c}: Since $f'(x)=10x-4$, we have $f'(2)= 10 \cdot 2 -4=20-4=16$ and
$f'(3)= 10 \cdot 3 -4=30-4=26$ and


{\bf 2} Using the famous formula, $(a+b)^3=a^3+3a^2b+3ab^2+b^3$ we can simplify $q$
$$
q= \frac{(v+2)^3 -8}{v}= \frac{v^3+ 3v^2 (2) + 3v (2^2)+ 2^3- 8}{v}=
\frac{v^3+ 6 v^2 + 12 v }{v}=
\frac{(v^2+ 6v + 12)v }{v}=v^2+6v+12 \quad .
$$

This is a nice polynomial in $v$ so to find the limit at a point, just {\bf plug-it in}

{\bf a.} $\lim_{ v \rightarrow 0} \, q = 0^2+6 \cdot 0 + 12 = 12$

{\bf b.} $\lim_{ v \rightarrow 2} \, q = 2^2+6 \cdot 2 + 12 = 4+12+12=28$

{\bf c.} $\lim_{ v \rightarrow a} \, q = a^2+6 \cdot a + 12 = 4+12+12=28$


{\bf 3.} We use the constant multiple rule (you can always take a constant out of the differentiation) and the famous rule
$$
\frac{d}{dx} x^n = n x^{n-1} \quad
$$

{\bf a.}
$$
\frac{d}{dx} \left ( -x^{-4}  \right ) =  -  \left ( \frac{d}{dx} x^{-4} \right ) =  - (-4) x^{-5} = 4x^{-5} \quad .
$$

{\bf b.}
$$
\frac{d}{du} ( a u^b )  =  a (\frac{d}{du} u^{b} =  a b \, u^{b-1}  \quad .
$$


{\bf c.}
$$
\frac{d}{dx} c x^2  =  c (\frac{d}{dx} x^{2} =  c (2x)= 2c \, x  \quad .
$$


{\bf d.}
$$
\frac{d}{dx} ( 7 x^{1/3} ) =  7 (\frac{d}{dx} x^{1/3} ) =  7 \cdot (1/3) x^{-2/3}= \frac{7}{3} x^{-\frac{2}{3}} \quad .
$$


{\bf 4.} The product rule is $(f(x)g(x))'=f'(x) g(x) + f(x) g'(x)$.

{\bf a.} $((9x^2-1)(3x+1))'= (9x^2-1)' (3x+1)+ (9x^2-1)(3x+1)'= (18x)(3x+1)+ (9x^2-1)3= 54x^2+18x +27x^2-3=81x^2 + 18x - 3$

{\bf b.} $((ax-b)(cx^2))'= (ax-b)' (cx^2)+ (ax-b)(cx^2)'= (a)(cx^2)+ (ax-b)(2cx)= ac \, x^2 + 2ac \, x^2 - 2bc \, x= 3ac \, x^2 -2bc\, x$

{\bf c.} 
$$
[(2-3x)(1+x)(x+2)]'= (2-3x)'(1+x)(x+2) +  (2-3x)(1+x)'(x+2) +  (2-3x)(1+x)(x+2)'=
(-3) (1+x)(x+2) +  (2-3x)(1)(x+2) +  (2-3x)(1+x)(1)
$$
$$
= -3(1+x)(x+2) +  (2-3x)(x+2) +  (2-3x)(1+x) \quad .
$$

{\bf d.} 
$$
[(x^2+3)x^{-1}]'= (x^2+3)'(x^{-1})+ (x^2+3)(x^{-1})'=
(2x)(x^{-1})+ (x^2+3)(-1)x^{-2} = 2-(1+ 3x^{-2})= 1- 3x^{-2} \quad .
$$

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

{\bf a.}
$$
((x^2+3)/x)'= (x+ 3/x)'= (x+ 3x^{-1})'= x'+ 3 (x^{-1})' = 1 + 3(-1)x^{-2}= 1- \frac{3}{x^2} \quad .
$$


{\bf b.}
$$
((x+7)/x)'= (1+ 7/x)'= (1+ 7x^{-1})'= 1'+ 7 (x^{-1})' = 0 + 7(-1)x^{-2}= -\frac{7}{x^2} \quad .
$$


The {\bf quotient rule} is:
$$
\left ( \frac{f(x)}{g(x)} \right )'=
\frac{f'(x) g(x) - f(x) g'(x)}{g(x)^2} \quad .
$$

{\bf c.}
$$
\left ( \frac{4x}{x+5} \right )'= \frac{(4x)' (x+5) - 4x (x+5)'}{(x+5)^2} 
= \frac{4 (x+5) - 4x (1)} {(x+5)^2} 
= \frac{4x+20 - 4x } {(x+5)^2} 
= \frac{20} {(x+5)^2}  \quad .
$$

{\bf d.}
$$
\left ( \frac{ax^2 + b}{cx+d} \right )'= \frac{(ax^2+b)' (cx+d) - (ax^2+b) (cx+d)'}{(cx+d)^2} 
=\frac{(2ax) (cx+d) - (ax^2+b)(c) }{(cx+d)^2} 
$$
$$
=\frac{2ac \, x^2 +2ad \, x - ac \, x^2  -bc}{(cx+d)^2} 
=\frac{ac \, x^2 \, + \, 2ad \, x \, - \, bc}{(cx+d)^2} 
\quad .
$$

{\bf 6.} The {\bf chain rule} says
$$
\frac{dw}{dx}= \frac{dw}{dy}  \cdot  \frac{dy}{dx}  \quad .
$$
Here
$$
\frac{dw}{dy} =  2ay \quad , \quad  \frac{dy}{dx} = 2bx \quad , \quad 
$$
Hence
$$
\frac{dw}{dx}=  (2ay) (2bx) =4ab xy \quad .
$$
{\bf Finally} we replace $y$ by what it stands for in terms of $x$, getting
$$
\frac{dw}{dx}= 4ab xy = 2abx(bx^2+cx)= 2ab x^2 (bx+c) \quad .
$$

{\bf 7.}
$$
((16x+3)^{-2})'=
(-2) ((16x+3)^{-3}) \cdot (16x +3)' =
(-2) ((16x+3)^{-3}) \cdot (16)= -32 (16x+3)^{-3} \quad .
$$

By the quotient rule:
$$
(\frac{1}{(16x+3)^2})'=
\frac{1'(16x+3)^2 - 1 ((16x+3)^2)' }{(16x+3)^4}=
\frac{0 \cdot (16x+3)^2 - 1 ((2)16x+3))(16) }{(16x+3)^4}=
\frac{-32}{(16x+3)^3} = -32 (16x+3)^{-3} \quad .
$$

The answers are identical because God (or Newton) said so!

{\bf 8.}

{\bf a.} When $x$ is {\bf VERY BIG} $-7x+5$ and $-1$ are {\bf insignifcant} compared to $2x^2$ and $x$ respectively,
hence it is possible to only retain the leading powers at the top and bottom.
$$
\lim_{x \rightarrow \infty} \frac{x-1}{2x^2-7x+5} =
\lim_{x \rightarrow \infty} \frac{x}{2x^2} =
\lim_{x \rightarrow \infty} \frac{1}{2x} =
\frac{1}{2 \infty} = 0 \quad.
$$

{\bf Ans. to 8a}: $0$.

{\b.} Since the top and bottom happen to be $0$ at $x=1$, we can use L'H\^opital's rule:
$$
\lim_{x \rightarrow 1} \frac{x-1}{2x^2-7x+5} =
\lim_{x \rightarrow 1} \frac{(x-1)'}{(2x^2-7x+5)'} =
\lim_{x \rightarrow 1} \frac{1}{4x-7} =
\frac{1}{4 \cdot 1 -7} = -\frac{1}{3} \quad .
$$

{\bf Ans. to 8b}: $-\frac{1}{3}$.

{\bf Comment:} We can also do it  only using algebra
$$
\lim_{x \rightarrow 1} \frac{x-1}{2x^2-7x+5} =
\lim_{x \rightarrow 1} \frac{x-1}{(x-1)(2x-5)} =
\lim_{x \rightarrow 1} \frac{1}{2x-5} =\frac{1}{2 \cdot 1-5} =  -\frac{1}{3} \quad .
$$

{\bf 9.} 

{\bf First way}: Using algebra, $f(x)=x+1$ when $x \neq 1$. 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.

{\bf Second Way}: $f(x)$ is continuous when $x \neq 1$ (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'H\^opital 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.

{\bf 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!

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