1. Differential Calculus
Concepts & Problems
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2. Matrix Properties
(An)-1 = (A-1)n (inverse (expt-matrix matrix n)) (expt-matrix (inverse matrix) n)) (AT )-1 = (A-1)T (AB)T = BTAT (AB)-1 = B-1A-1 det A = 1 / det A-1 Determinants can be calculated using co-factors and minors. Only square matrices have determinants.
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3. Inverse Matrix
AA-1 = I, the Identity matrix. Find the inverse matrix of #2A( (1 2) (3 8) ) 1a + 3b = 1 2a + 8b = 0 1c + 3d = 0 2c + 8d = 1 (solve '((1 3 1)(2 8 0)))  (4 -1) (solve '((1 3 0) (2 8 1)))  (-3/2 1/2)
(4 -1)
(-3/2 1/2)
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4. Determinants
|A| = |AT| has determinant has determinant =
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5. Co-factors
Evaluate (7 – 12) = 5 (det #2A ((1 2 3)(4 6 7)(0 1 0)))  5 Find the inverse,
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6. Eigenvalues Find eigenvalues for matrix [-5 9; -6 10]
(eigenvalues #2A((-5 9)(-6 10)))  (1, 4)
t t – 10 = t2 -5t + 4 = 0 => (t – 1)(t – 4) = 0
Eigenvectors are 6 6 or (1, 1), (9, -6) or (3, -2).
Note orthogonality of the two vectors =
(M* (inverse #2A((1 3)(1 2))) #2A((-5 9)(-6 10)) #2A((1 3)(1 2)))
#2A((4 0)(0 1))
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7. Cryptography
A = Message: SEND HELP = GOKI0ATG = recovered message
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8. Getting There From Here
Limits and Continuity
Getting There From Here
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9. Limit of a Sequence
1 3/2 5/3 7/4 9/5 11/6 … 2 – 1/n, … limit  2 but the sequence does not contain 2. On the other hand the sequence 1 ½ 1 ¾ 1 5/6 1 6/7 …has 1 as a limit and contains 1
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10. Limit of a Function
1 3/2 5/3 7/4 9/5 11/6 … 2 – 1/n  2 As x  2, f(x) = x2  4 1 9/4 25/9 49/16 81/25 121/36  4 Find limits of 1 ½ 1/3 ¼ 1/5 … 1/n /3 7/2 17/5 … 3 + 2/n ½ ¼ 1/8 1/16 1/32 …
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11. /2
Some famous limits
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12. Limit
Find the limits
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13. Limit of a Series
1 + ½ + ¼ + 1/8 + 1/16 + … sum of terms of a sequence (+ 1 1/2 1/4 1/8 1/16 1/32 1/64 1/128 1/256 1/512 1/1024)  1.999
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14. Finding a Limit
Find
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15. L'Hospital's Rule
Applies to Indeterminate Forms
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16. e
Evaluate Use L'Hospital's Rule on [ln(1 + x)]/x
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17. Continuity
f is continuous at x = a if
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18. Where Was the Farmer?
When he jumped off the roof? On the roof. That's where he was before he jumped. Well then in the air. That's where he was after he jumped. Where was he when he jumped?
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19. The Slope of a line passing through a curve
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20. The Slope of a line tangent at a point on the curve
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21. What is the Derivative?
The derivative is one of the two central concepts of calculus.
The derivative gives the slope of the line tangent to a function at a point. In this way, derivatives can be used to determine many geometrical properties of the function such as concavity or convexity.
The derivative provides a mathematical formulation of the instantaneous rate of change; it measures the rate at which the function's value changes as the function's argument changes.
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22. Let’s compute a derivative!
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23. The Notation All of the following are equivalent when y = f(x):
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24. Three Rules to live by
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25. Example Find the equation of the line tangent to
y = f(x) = 2x2 + 2x + 3 at the point (1,7)
f’(x) = 4x + 2 slope function
f’(1) = 6
Using the point-slope form of a straight line: y – 7 = 6(x – 1) or y = 6x + 1
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26. The Chain Rule
y=f[u(x)]
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27. Alternate approach
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28. The top 5 Table of Derivatives
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29. Table of Derivatives – the next 5
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30. Another Example
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31. and some more…
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32. x2, 2x and xx
y = x2 => y' = 2x y = 2x => ln y = x ln 2 => y'/y = ln 2 => y' = ln 2 * 2x y = xx => ln y = x ln x => y'/y = x * 1/x + ln x = 1 + ln x y' = xx(1 + ln x) Lim (2 + h)5 – 32 h  0 h
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33. Our First Application of Derivatives
Analyzing Functions
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34. Some Well Known Facts
Given f(x) is continuous and differentiable in the interval (a,b), then
If f’(x) = 0, x  (a, b), then f(x) is constant
If f’(x) > 0, x  (a, b), then f(x) is strictly increasing
If f’(x) < 0, x  (a, b), then f(x) is strictly decreasing + + -
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35. Convex and Concave Functions
If f’(x) is increasing, x  (a, b), f(x) is convex in the interval (a, b) If f’(x) is decreasing, x  (a, b),f(x) is concave in the interval (a, b)
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36. Example of Some Well Known Facts
f(x) = x3 – 2x + 1 and f’(x) = 3x2 – 2
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37. Maxima and Minima
Graph y = 3x2 – 12x Find where increasing. Graph y'.
y' = 6x – 12  for x > 2 => increasing 2
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38. Concavity y = 3x4 – 7x2 – 2; y' = 12x3 – 14x; y'' = 36x2 - 14
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39. Our Second Application of the Derivative
Finding Roots…
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40. Newton-Raphson Method
Problem: solve for x: f(x) = 0 when f(x) is not linear or quadratic.
Assume f(x) is continuous on the interval [a, b] and f(a) and f(b) have opposite sign, then the equation f(x) = 0 has at least one real root between a and b.
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41. Newton-Rhapson Method
Newton Iteration
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42. Newton-Raphson Method -2
f(x1)
x x1
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43. Newton-Raphson Method - 3
To find the solution to f(x) = 0
Recursion formula
Repeat the above in order to obtain a better approximation
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44. Newton-Raphson Method - 4
Find the root of f(x) = x4 – 4x + 1 = 0 in the interval (0,1)
f(0) = 1 and f(1) = -2 (a sign change, therefore at least one root)
f’(x) = 4x3 - 4
(quartic ) 
( d d0 #C( d d0) #C( d d0))
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45. Higher Order Derivatives
First derivative y’
f’(x)
Second derivative
y’’
f’’(x)
Third derivative
y’’’
f’’’(x)
Fourth derivative
Y(4)
f(4)
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46. An Example f(x) = 6x3 – 12x2 + 6x – 2 f’(x) = 18x2 – 24x + 6
f(4) = f(5) = … = 0
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47. Proceeding from a single variable function to a 2 variable function.
Partial Derivatives
Proceeding from a single variable function to a 2 variable function.
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48. Partial Derivatives
To find fx(x,y), treat y as a constant and differentiate f with respect to x
in the usual way.
To find fy(x,y) treat, x as a constant and differentiate f with respect to y
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49. Example Problems
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50. More Notation
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51. Partials
1st partials
2nd partials
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52. Partial Differentiation
y = f(x, y, z, t, …) is differentiated with respect to one of these variables, the other variables are held constant.
Let w = f(x, z) = x2 + xz2.
Then Wx =
Wz =
Consider right circular cylinder V(d, h) = d2h/4;
Vd = dh/2 and
Vh = d2/4; the area
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53. Unconstrained Optimization
Find the area bounded by the curve y = f(x) = x3 - 6x2 + 9x + 1 and the x-axis between the curve's x-extrema. y'(x) = 3x2 –12x + 9 = 0 when x = 1 or x = 3 pt of inflection y''(x) = 6x – 12; x = 2 is Pt-Inflection y''(1) < 0 => relative max at(1,5); y''(3) > 0 => relative min at (3,1) = 6
(U-poly-eval '( ) 1)
syms x; f = x^3 - 6*x^2 + 9*x + 1; fx = diff(f); roots([ ]);ans = ; fxx1 = -6.00
fxx2 = subs(diff(fx),3) =fxx2 = 6.00; subs(f,1) = 5.00; subs(f,3) = 1.00
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54. Cost to Construct
A box has base length 3 times the width. The cost to construct is $10/ ft2 for the top and bottom and $6/ft2 for the sides. The volume must be 50 ft3. Find the dimensions to minimize the cost to construct. C = 2(10)(3w2) + 2(6)(hw) + 2(6)(3hw) = 60w2 + 48hw; 50 = 3w2h => h = 50/3w2 C = 60w /w C' = 120w - 800/w2 = 0 when 12w3 = 80 => w = 1.88 C" = /w > 0 => relative minimum. h
l=3w w
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55. Z = f(x, y) = x3 + y3 + 3xy
Zx = 3x2 + 3y = 0; Zy = 3y2 + 3x = 0; (0, 0); (-1 -1) x2 + y2 + x + y = 0; or (x + ½)2 + (y + ½)2 = ½ Zxx = 6x; Zxy = 3 = Zyx Circle centered at (½, ½) with r = 1/21/2 Rules: D & Zxx > 0 => rel min; D > 0 and Zxx < 0 rel max D < 0 => saddle point and D = 0 => no info D = At (-1 -1), D = 25 and Zxx < 0 => rel max; Z(-1, -1) = 1 At (0, 0), D = -9 => saddle point at (0, 0, 0)
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56. f(x, y) = x2 + y2
fx = 2x; fy = 2y (0, 0) fxx = 2; fyy = 2 D = = 4 > 0 and fxx and fyy > 0 => relative minimum
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57. Maxima and Minima
f(x, y) = 5x2 + 10y2 + 12xy – 4x – 6y + 1 fx = 10x + 12y – 4 = 0; fy = 20y + 12x – 6 = 0 (solve '(( )( )))  (1/7 3/14) f(1/7, 3/14) = -1/14 fxx = 10 fxy = 12 fyy = 20 => (1/7, 3/14, -1/14) is a relative minimum.
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58. f(x, y) = 3x2y + x2 - 6x – 3y - 2
fx = 6xy + 2x – 6; fy = 3x2 - 3 fxy = 6x; fyx = 6x fxx = 6y + 2; fyy = 0 Substitute x = 6/(6y + 2) to get 36/(6y + 2)2 = 1 or 36 = 36y2 + 24y + 4 or (quadratic ) (2/3 -4/3) -8 4 D = -16 => saddle points; no rel max or min. 4 0
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59. Cross Product
Let vector u = 2i – 3j + k and v = i + j – 2k. Then Note u  (u x v) = = 0 and v  (u x v) = = 0 => to (u x v) u x v = |u||v| sin  i, j and k form a right-handed triple u  v = |u||v| cos  in that i x i = j x j = k x k = 0 i x j = k; j x k = i, k x i = j
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60. Applied Problems In Extrema
1. Divide 120 into 2 parts such that the product of one times the square of the other is a maximum. f(x) = x2(120 – x) = on [0, 120]. f'(x) = 240x – 3x2 = 0 when x = 0, x = 80 f"(x) = 240 – 6x f"(0) = 240 > 0 => relative minimum f"(80) = -240 < 0 => relative maximum f(0) = 0, f(120) = 0 => f(80) = 256,000 is max point.
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61. 2. Show that the curve y = x3 – 8 has no maximum or minimum.
f '(x) = 3x2 = 0 at x = 0
When x is less or greater than zero, the slope is positive and thus constantly increasing.
f "(0) = 0 => point of inflection.
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62. (dD2/dt)' = 650 => relative minimum
3. Ship B is 65 miles due east of ship A and is sailing due west at 10 mph while A is sailing due south at 15 mph. Find their closest distance to each other if they continue.
65 – 10t ,
A B *131/2
15t
D2 = (15t)2 + (65 – 10t)2
= 225t – 1300t + 100t2
dD2/dt = 650t – 1300 = 0 when t = 2
(dD2/dt)' = 650 => relative minimum
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63. Optimization
Given a length of L units to create a square and a circle, find the dimensions to maximize the area. Of all planar areas the circle has minimum perimeter. x
L - x
x/4
C = 2r = L – x
C2 = 42r2
f(x) = 0 x L
f ‘(x) = f(0) = => all for the circle; absolute maximum
f(L) = => all for the square
f ‘(x) =
f() = ; absolute minimum
f(x) = x2/16 + (L – x)2/4, 0  x  L; L2/4; L2 / f'(x) = x/8 + (L – x)/2 = 0 at x = 4L/(+ 4)
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64. Z = f(x, y) = x3 + y3 + 3xy
Zx = 3x2 + 3y = 0; Zy = 3y2 + 3x = 0; Test: (0, 0); (-1 -1) x2 + y2 + x + y = 0; or (x + ½)2 + (y + ½)2 = ½ Zxx = 6x; Zxy = 3 = Zyx; Zyy = 6y D = At (-1 -1), D = 36 – 9 = 25; Zxx < 0 => rel max; Z(-1, -1) = 1 At (0, 0), D = -9 => saddle point at (0, 0, 0)
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65. Exercises
The sum of 2 positive numbers is 20. Find if product is a maximum. (10, 10) Find if sum of their squares is a minimum. (10, 10) Find if product of the square of one and the cube other the other is a maximum. (12, 8)
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66. Exercises
The product of two positive numbers is 16. a) Find the numbers if the sum is least. (4, 4) b) If the sum of one and square of other is least. (8, 2) Find equation of line through the point (3, 4) which cuts the first quadrant a triangle of minimum area. 4x + 3y – 24 = 0 (3, 4)
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67. Row Your Boat
You are 5 miles from shore and wish to land 6 miles along the shore in minimum time. You can row at 2 mph and walk at 4 mph. Where should you land? P 5 (25 + x2)1/2 C A x 6 - x
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68. Tangents and Normals
Problems
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69. 2x – xy' - y + 2yy' = 0 => y' = (y – 2x)/(2y – x)
1. Find the points of tangency of horizontal and vertical tangents to x2 –xy + y2 = 27.
2x – xy' - y + 2yy' = 0 => y' = (y – 2x)/(2y – x)
Horizontal: y – 2x = 0 or y = 2x and
x2 –xy + y2 = 27 => x2 – x(2x) + 4x2 = 27 => 3x2 = 27 and x2 = 9,
Points (3, 6) and (-3, -6)
Vertical: 2y – x = 0 or x = 2y and x2 –xy + y2 = 27
4y2 -2y2 + y2 = 27 => 3y2 = 27; (6, 3) and (-6, -3)
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70. 2. Find the equation of the tangent and the normal to
x2 + 3xy + y2 = 5 at (1, 1).
2x + 3xy' + 3y + 2yy' = 0 => y' = -(2x + 3y)/(3x + 2y)
The slope of the tangent at (1, 1) is -1.
Tangent: (y – 1) = -1(x – 1) or x + y = 2
Normal: (y – 1) = 1(x – 1) or x – y = 0
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71. 2yy' = 4p or y' = 2p/y. Let(xo, yo) be the point of tangency. Then
3. Show that the equation of the tangent of slope m  0 to the parabola y2 = 4px is y = mx + p/m.
2yy' = 4p or y' = 2p/y. Let(xo, yo) be the point of tangency. Then
y02 = 4px0 and m = y'(y0) = 2p/y0
m = 2p/y => y0 = 2p/m and x0 = (4p2/m2)/4p = p/m2
(y – 2p/m = m(x – p/m2)
y = mx + p/m
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72. 4. Find extrema of y = (x – 2)2/3.
y' = 2(x – 2)-1/3 /3 = 2/3(x – 2)1/3.
Critical value at x = 2
Use first derivative test
As x  2+ positive slope;
As x  2- negative slope indicating relative minimum 2
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73. Related Rates
Air is escaping a spherical balloon at the rate of 2 ft3/min/ How fast is the surface area shrinking when the radius is 12 ft? Vsphere = 4r3/3 and Ssphere = 4r2 dV/dt = 4r2 * dr/dt; dS/dt = 8r * dr/dt dS/dt /dV/dt = dS/dV = 2/r = 2/12 = 1/6 ft2/ft3 dS/dV * dV/dt = dS/dt = 1/6 * -2 = -1/3 ft2/min ft2/ft3 * ft3/min = ft2 / min
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74. Exponential Extrema
Find extrema of f(x) = x2ex. f'(x) = x2ex + 2xe = xex(x + 2) = 0 for x = 0, -2 (0, 0) is relative min and (-2, 4e-2) is relative max. f"(x) = 2ex + 4xex + x2e = ex(2 + 4x + x2) (quadratic 1 4 2)  ( , ) (-2  21/2) are points of inflection.
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75. Rate of Change
Show that the rate of change of the area of a circle with respect to its circumference is its radius. A = R2 = 2R2/2 = CR/2 = C2/4 => dA/dC = 2C/4 = R C = 2R dA/dC = dA/dR * dR/dC = 2R * 1/2 = R
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76. Motion
A ball thrown vertically upwards has position given by s = 218t – 16t2 . Find the times when the velocity is 64 ft/sec. s' = ds/dt = v = 218 – 32t = 64 when t = 154/32 = sec.
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77. Rate of Change
The volume of a ball is V = d3/6. Find the rate of change of the volume with respect to the diameter when the diameter is 2 ft. dV/dd = 3d2/6 = ½ *  * 4 = 2.
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78. 536-56
Given s = 9/(2t 2 + 3), find v at t = 1. s' = v = -9(4t)/(2t 2 + 3)2 v(1) = -36/25 ft/sec
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79. Logarithmic Differentiation
y = (2x - 5)3 / x2(x2 + 1)1/4 ln y = 3 ln (2x – 5) – 2 Ln x – ¼ ln(x2 + 1) y'/y = 6/(2x – 5) – 2/x – ¼ (2x/x2 + 1) y' = (2x - 5)3 / x2(x2 + 1)1/4 * 6/(2x – 5) – 2/x – ¼ (2x/x2 + 1)
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80. Logarithmic Differentiation
y = (1 + ex) ln x ln y = (ln x) ln (1 + ex) y'/y = ln x * ex/(1 + ex) + ln (1 + ex) (1/x) y' = (1 + ex) ln x *[ ln x * ex/(1 + ex) + ln (1 + ex) (1/x)]
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81. Implicit Differentiation
Given ey = (y + 1)ex , find dy/dx. ey y' = ex(y + 1) + y'ex y'(ey - ex ) = ex (y + 1) y' = ex(y + 1)/(ey - ex ) = ey / (ey – ey / (y+ 1) = 1/[1 – 1/(y + 1) = (y + 1)/y
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82. Implicit Differentiation
Find y' at (1,1) given that x2y - xy2 + x2 + y2 = 0. 2xy + x2y' -2xyy' –y2 + 2x + 2yy' = 0 y'(x2 – 2xy + 2y) = y2 – 2xy -2x y' = -3
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83. Tangency Between Curves
Find the angle of intersection of curves y2 = 4x and 2x2 = 12 – 5y Pts of intersection (1, 2) and (4, -4) 2yy' = 4 => y' = 2/y; y' = -4x/5 At (1, 2), m1 = 1; m2 = -4/5 At (4,-4), m1 = -1/2; m2 = -16/5 At P1, tan  = (m1 – m2)/(1 + m1m2) (atan 9) = 1.46 radians = (1 + 4/5)/(1 – 4/5) = 9 =>  = 83 40' At P2 tan  = (-1/2 + 16/5)/(1 + 8/5) = =>  = 46 5'
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84. Asymptotes
f(x) = (x2 – 4x) / (x2 - 4x + 3) Horizontal : lim as x   is 1 Vertical: (quadratic 1 –4 3)  1, 3
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85. Vector Analysis
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86. Vector Derivative
Geometric Interpretation r(t) = xi + yj + zk dr/ds = T, a unit vector along tangent dr/dt = velocity d2r/dt2 = acceleration
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87. Gradient, Divergence, Curl
Let (x,y,z) be a scalar function and A(x,y,z) a vector function
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88. Gradient, Divergence, Curl
Let (x, y, z) = x2yz3, A = xzi –y2j + 2x2yk Then  = 2xyz3i + x2z3j + 2x2yz2k, perpendicular to    A = z -2y  x A = = 2x2i + (x – 4xy)j
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89. F(x, y) = y3 + x2 - 6xy + 3x + 6y - 7
Fx = 2x – 6y + 3 = 0; Fy = 3y2 - 6x + 6 = 0 2(3y2 + 6)/6 – 6y + 3 = 0 (quadratic )  y-values (1, 5); x-values (3/2, 27/2) Fxx = 2; Fxy = -6 Fyx = -6 Fyy = 6y D = 12y – 36 For y = 1, D < 0 => saddle point at (3/2, 1) y = 27/2, D > 0 & Fxx > 0 => relative minimum
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90. Conics
locus of points in plane whose distance from a fixed point (focus) and a from a given straight line (directrix) have constant ratio. e = r/a e = 0 => circle 0 < e < 1 => ellipse a r e = 1 => parabola e > 1 => hyperbola a r
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