# Derivative and Calculus Book

Topics: Derivative, Integral, Convex function Pages: 9 (2131 words) Published: May 12, 2013
Section 1.2 (Page 87) (Calculus Book): 14, 23, 26, 29, 30, 31, and 32 14. limt→1t3+t2-5t+3t3-3t+2
=limt→1t3+t2-5t+3t3-t2+t2-t-2t+2
=limt→1t3-t2+2t2-2t-3t+3t2t-1+tt-1-2t-1
=limt→1t2t-1+2tt-1-3t-1t2+t-2t-1
=limt→1t2t-1+2tt-1-3t-1t-2t-1
=limt→1t-1t2+2t-3t-2t-1
=limt→1t2+3t-t-3t2+2t-t-2
=limt→1tt+3-1t+3tt+2-1t+2
=limt→1t+3t-1t+2t-1
=limt→1t+3t+2=1+31+2=43
23 limy→6y+6y2-36=limy→6y+6y+6y-6
⟹limy→61y-6=16-6=10=undefined ∴limit doesn't exist
26 limx→43-xx2-2x-8=limx→43-xx2-4x+2x-8=limx→4 3-xxx-4+2x-4 =lim x→43-xx-4x+2=3-44-44+2=-10×6=10
=undefined; ∴Limit does not exist.
29 limx→9x-9x-3=limx→9x2-32x-3=limx→9x-3x+3x-3
=limx→9x+3=9+3=3+3=6
30 lim y→44-y2-y=lim y→422-y22-y=lim y→42-y2+y2-y
=limy→42+y=2+4=2+2=4
31 fx=x-1, &x≤33x-7, &x>3
a. limx→3- fx=x-1=3-1=2
b. limx→3+ fx=3x-7=3×3-7=2
c. limx→3 fx=x-1=3-1=2
32 gt=t-2, t<0t2, 0≤t≤22t, t>2 a. limt→0 gt
As ‘t’ approaches to ‘0’ from the left side of the number line the applicable function is limt→0- gt=t-2=0-2=-2 As ‘t’ approaches to ‘0’ from the right side of number line the applicable function is limt→0+ gt=t2=02 From this we can conclude that at t=0 limit does not exist.

b. limt→1 gt
As ‘t’ approaches to ‘1’ from the both side of the number line the applicable function is limt→1 gt=t2=12=1 c. limt→2 gt
As ‘t’ approaches to ‘2’ from the left side of the number line the applicable function is limt→2- gt=t2=22=4 As ‘t’ approaches to ‘2’ from the left side of the number line the applicable function is limt→2+gt=2t=2.2=4 As the one-sided limits are equal, we can conclude that limt→2gt=4 Section 1.3 (Page 97) (Calculus Book): 13, 17, 23, 27, 28, 29, and 30 13 limx→∞3x+12x-5=limx→∞3x+1xlimx→∞2x-5x=limx→∞3+1xlimx→∞2-5x=3+0(2-0)=32 17 limx→-∞x-2x2+2x+1

=limx→-∞x-2x2limx→-∞ (x2+2x+1)/x2
=limx→-∞ 1x -2limx→-∞ 1x2 limx→-∞ 1+2 limx→-∞ 1x +limx→-∞ 1x2=0)(1+0+0)=0 23limx→-∞32+3x-5x21+8x2=limx→-∞32+3x-5x21+8x2 x2x2
=3limx→-∞ 2+3x-5x2x2 limx→-∞ 1+8x2x2
=3limx→-∞ 2x2+3limx→-∞ 1x-limx→-∞5limx→-∞1x2+limx→-∞ 8 =30+3.0 -50+ 8 =-5132313=-5132
27limy→-∞2-y7+6y2
=limy→-∞2-yylimy→-∞7+6y2y=lim y→-∞2-y-ylimy→-∞7+6y2y2 =limy→-∞2-y+1limy→-∞7y2+6y2y2=-2limy→-∞1y+limy→-∞1 7 limy→-∞1y2+limy→-∞6=0+17.0+6=16 28limy→+∞2-y7+6y2
=limy→+∞2-yylimy→+∞7+6y2y
=lim y→+∞2-yylimy→+∞7+6y2y2
=limy→+∞2y-1limy→+∞7y2+6y2y2=0-10+6=-16
29limx→-∞3x4+xx2-8
=limx→-∞3x4+xx2x2-8x2
=limx→-∞3x4+xx4x2-8x2
=limx→-∞3x4x4+xx4x2x2 – 8x2
=limx→-∞3+1x31 - 8x2
= limx→-∞3 +limx→-∞ 1x3 limx→-∞ 1x2 -8 limx→-∞ 1x2 =3.1-01-8.0=3
30limx→+∞3x4+xx2-8
=limx→+∞3x4+xx2x2-8x2
=limx→+∞3x4+xx4x2-8x2
=limx→-∞3x4x4+xx4x2x2 – 8x2
=limx→-∞3+1x31 – 8x2
= limx→-∞3 +limx→-∞ 1x3 limx→-∞ 1x2 –8 limx→-∞ 1x2 =3.1-01-8.0=3

Section 2.1 (Page 141) (Calculus Book): 15, 16
A function y = f(x) and an x-value x0 are given.
(a) Find a formula for the slope of the tangent line to the graph of f at a general point x = x0. (b) Use the formula obtained in part (a) to find the slope of the tangent line for the given value of x0. 15 fx=x2-1 ;x0=-1

A. mtan = limx→x0fx- fx0x-x0
= limx→x0x2-1-x02-1x-x0
=limx→x0x2-x02x-x0
=limx→x0x+x0
= x0+x0
=2x0
B. mtan=2x0=2-1=-2
16 fx=x2+3x+2 ;x0=2
a. mtan = limx→x0fx- fx0x-x0
b. = limx→x0x2+3x+2-x02+3x0+2x-x0
=limx→x0x2-x02+3x-x0+0x-x0
=limx→x0x+x0x-x0+3x-x0x-x0
=limx→x0x+ x0+3
=x0+ x0+3
=2x0+3
c. mtan=2x0+3=22+3=7

Section 2.2 (Page 152) (Calculus Book): 12,14,17,20
According to definition 2.2.1, the function f' defined by the formula f'x=limh→0fx+h-fxh is called the derivative of 'f' with respect to 'x'. The domain of f'consists of all x in the domain of f for which the limit exists. 12 fx=2x3+1;a=-1

f'-1=limh→0f-1+h-f-1h
=limh→02-1+h3+1-2-13+1h
= limh→06-6h+2h2
=6
Now, f-1=2-13+1=-2+1=-1
And f'-1=6
the tangent line, y-y1=m x-x1
⟹y--1=f'xx--1
⟹y+1=6x+1
⟹y=6x+6-1
∴y=6x+5
14 fx=2x+1;a=4
f'4=limh→0f4+h-f4h
⟹ limh→024+h+1-24+1h
⟹ limh→09+2h-9h
⟹...