Chapter 2
22 mai 2017 be seen from the second derivative (if it exists). 6. Page 7. DMM summer 2017 ... Calculate the gradient of fA
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15 nov. 2012 determinant derivative of inverse matrix
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Gradients of Inner Products
b − Ax2. 2. = (b − Ax)T (b − Ax). = bT b − (Ax)T b − bT Ax + xT AT Ax. = bT b − 2bT Ax + xT AT Ax Because mixed second partial derivatives satisfy.
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determinant derivative of inverse matrix
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Techniques of Integration
apparent that the function you wish to integrate is a derivative in some EXAMPLE 8.1.2 Evaluate ∫ sin(ax + b) dx assuming that a and b are constants ...
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Week 3 Quiz: Differential Calculus: The Derivative and Rules of
Question 2: Find limx→2f(x): f(x) = 1776. (A) +∞. (B) 1770. (C) −∞. (D) Does not exist! (E) None of the above. Answer: (E) The limit of any constant
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Assignment 2 — Solutions
If a1b2 = a2b1 show that this equation reduces to the form y′ = g(ax + by). Solution Substituting a2 = λa1 and b2 = λb1 into the equation yields:.
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To see that action I will write b1
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1 Theory of convex functions
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Order and Degree and Formation of Partial Differential Equations
When a differential equation contains one or more partial derivatives of an (viii) z = ax e' +. 1. 2. Sol. (1) We are given z = (2x + a) (2 y + b).
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Week 3 Quiz: Dierential Calculus: The Derivative and Rules of Dierentiation SGPE Summer School 2016
Limits
Question 1:Find limx!3f(x):
f(x) =x29x3 (A) +1 (B) -6 (C) 6 (D) Does not exist! (E) None of the above Answer:(C) Note the the functionf(x) =x29x3=(x3)(x+3)x3=x+ 3 is actually a line. However it is important to note the this function isundenedatx= 3. Why?x= 3 requires dividing by zero (which is inadmissible). Asxapproaches 3 from below and from above, the value of the functionf(x) approachesf(3) = 6. Thus the limit limx!3f(x) = 6.Question 2:Find limx!2f(x):
f(x) = 1776 (A) +1 (B) 1770 (C)1 (D) Does not exist! (E) None of the above Answer:(E) The limit of any constant function at any point, sayf(x) =C, whereCis an arbitrary constant, is simplyC. Thus the correct answer is limx!2f(x) = 1776.Question 3:Find limx!4f(x):
f(x) =ax2+bx+c (A) +1 (B) 16a + 4b + c (C)1 (D) Does not exist! (E) None of the above 1Answer:(B) Applying the rules of limits:
lim x!4ax2+bx+c= limx!4ax2+ limx!4bx+ limx!4c =a[limx!4x]2+blimx!4x+c = 16a+ 4b+cQuestion 4:Find the limits in each case:
(i) lim x!0x 2jxj (ii) lim x!32x+34x9 (iii) lim x!6x23xx+3
Answer:(i) limx!0x
2jxj= limx!0(jxj)2jxj= limx!0jxj= 0
(ii) limx!32x+34x9=23+3439= 3 (iii) limx!6x23xx+3=62366+3
= 2 Question 5:Show that limx!0sinx= 0 (Hint:xsinxxfor allx0.) Answer:Given hint and squeeze theorem we have limx!0x= 0limx!0sinx0 = limx!0xhence, lim x to0sinx= 0Question 6:Show that limx!0xsin(1x
) = 0 Answer:Note rst that for any real numbertwe have1sint1 so1sin(1x )1. Therefore, xxsin(1x )xand by squeeze theorem limx!0xsin1x = 0.Continuity and Dierentiability
Question 7:Which of the following functions areNOTeverywhere continuous: (A)f(x) =x24x+2 (B)f(x) = (x+ 3)4 (C)f(x) = 1066 (D)f(x) =mx+b (E) None of the above Answer:(A) Remember that, informally at least, acontinuousfunction is one in which there are nobreaks its curve. A continuous function can be drawn without lifting your pencil from the paper. More
formally, a functionf(x) iscontinuousat the pointx=aif and only if:1.f(x) is dened at the pointx=a,
2. the limit lim
x!af(x) exists,3. lim
x!af(x) =f(a) The functionf(x) =x24x+2is not everywhere continuous because the function is not dened at the point x=2. It is worth noting that limx!2f(x) does in fact exist!The existence of a limit at a point does not guarantee that the function is continuous at that point! 2 Question 8:Which of the following functions are continuous: (A)f(x) =jxj (B)f(x) =3x <4 12 x+ 3x4 (C)f(x) =1x (D)f(x) =lnx x <0 0x= 0 (E) None of the above Answer:(A) The absolute value functionf(x) =jxjis dened as: f(x) =x x0 x x <0 Does this function satisfy the requirements for continuity? Yes! The critical point to check isx= 0. Note that the function is dened atx= 0; the limx!0f(x) exists; and that limx!0f(x) = 0 =f(0). Question 9:Which of the following functions areNOTdierentiable: (A)f(x) =jxj (B)f(x) = (x+ 3)4 (C)f(x) = 1066 (D)f(x) =mx+b (E) None of the above Answer:(A) Remember that continuity is anecessarycondition for dierentiability (i.e., every dier- entiable function is continuous), but continuity is not asucientcondition to ensure dierentiability(i.e., not every continuous function is dierentiable). Case in point isf(x) =jxj. This function is in
fact continuous (see previous question). It is not however dierentiable at the pointx= 0. Why? The pointx= 0 is a cusp (or kink). There are an innite number of lines that could be tangent to the functionf(x) =jxjat the pointx= 0, and thus the derivative off(x) would have an innite number of possible values.Question 10:Is function
f(x) =0 :x= 0 xsin(1=x) :x6= 0 continuous at point 0?Answer:Note thatfis continuous at a pointaif
lim x!af(x) =f(limx!ax):In this case, we takea= 0 and
lim x!0f(x) = limx!0xsin(1=x) = 0 by question 6. Moreover, f(limx!0x) =f(0) = 0 thus,fis continuous at 0. 3Derivatives
Question 11:Find the derivative of the following function: f(x) = 1963 (A) +1 (B) 1963 (C)1 (D) 0 (E) None of the above Answer:(D) The derivative of a constant function is always zero. Question 12:Find the derivative of the following function: f(x) =x2+ 6x+ 9 (A)f0(x) = 2x+ 6 + 9 (B)f0(x) =x2+ 6 (C)f0(x) = 2x+ 6 (D)f0(x) = 2x (E) None of the aboveAnswer:(C) Remember that 1) the derivative of a sum of functions is simply the sum of the derivatives
of each of the functions, and 2) the power rule for derivatives says that iff(x) =kxn, thenf0(x) = nkx n1. Thusf0(x) = 2x21+ 6x11+ 0 = 2x+ 6. Question 13:Find the derivative of the following function: f(x) =x12 (A)f0(x) =12 px (B)f0(x) =1px (C)f0(x) =12 px (D)f0(x) =px (E) None of the above Answer:(C) Remember that the power rule for derivatives works with fractional exponents as well!Thusf0(x) =12
x12 1=12 x12 =12 px Question 14:Find the derivative of the following function: f(x) = 5x2(x+ 47) (A)f0(x) = 15x2+ 470x (B)f0(x) = 5x2+ 470x (C)f0(x) = 10x (D)f0(x) = 15x2470x 4 (E) None of the above Answer:(A) Ideally, you would solve this problem by applying the product rule. Setg(x) = 5x2and h(x) = (x+ 47), thenf(x) =g(x)h(x). Apply the product rule: f0(x) =g0(x)h(x) +g(x)h0(x)
= 10x(x+ 47) + 5x2(1) = 10x2+ 470x+ 5x2 = 15x2+ 470x Question 15:Find the derivative of the following function: f(x) =5x2x+ 47 (A)f0(x) =5x2470x(x+47)2 (B)f0(x) =10x2+470x(x+47) (C)f0(x) = 10x (D)f0(x) =5x2+470(x+47)2 (E) None of the above Answer:(A) Ideally, you would solve this problem by applying the quotient rule. Setg(x) = 5x2and h(x) = (x+ 47), thenf(x) =g(x)h(x). Apply the quotient rule: f0(x) =g0(x)h(x)g(x)h0(x)h(x)2
10x(x+ 47)5x2(1)(x+ 47)2
10x2+ 470x5x2(x+ 47)2
5x2+ 470x(x+ 47)2
Question 16:Find the derivative of the following function: f(x) = 5(x+ 47)2 (A)f0(x) = 15x2+ 470x (B)f0(x) = 10x470 Week 3 Quiz: Dierential Calculus: The Derivative and Rules of DierentiationSGPE Summer School 2016
Limits
Question 1:Find limx!3f(x):
f(x) =x29x3 (A) +1 (B) -6 (C) 6 (D) Does not exist! (E) None of the above Answer:(C) Note the the functionf(x) =x29x3=(x3)(x+3)x3=x+ 3 is actually a line. However it is important to note the this function isundenedatx= 3. Why?x= 3 requires dividing by zero (which is inadmissible). Asxapproaches 3 from below and from above, the value of the functionf(x) approachesf(3) = 6. Thus the limit limx!3f(x) = 6.Question 2:Find limx!2f(x):
f(x) = 1776 (A) +1 (B) 1770 (C)1 (D) Does not exist! (E) None of the above Answer:(E) The limit of any constant function at any point, sayf(x) =C, whereCis an arbitrary constant, is simplyC. Thus the correct answer is limx!2f(x) = 1776.Question 3:Find limx!4f(x):
f(x) =ax2+bx+c (A) +1 (B) 16a + 4b + c (C)1 (D) Does not exist! (E) None of the above 1Answer:(B) Applying the rules of limits:
lim x!4ax2+bx+c= limx!4ax2+ limx!4bx+ limx!4c =a[limx!4x]2+blimx!4x+c = 16a+ 4b+cQuestion 4:Find the limits in each case:
(i) lim x!0x 2jxj (ii) lim x!32x+34x9 (iii) lim x!6x23xx+3
Answer:(i) limx!0x
2jxj= limx!0(jxj)2jxj= limx!0jxj= 0
(ii) limx!32x+34x9=23+3439= 3 (iii) limx!6x23xx+3=62366+3
= 2 Question 5:Show that limx!0sinx= 0 (Hint:xsinxxfor allx0.) Answer:Given hint and squeeze theorem we have limx!0x= 0limx!0sinx0 = limx!0xhence, lim x to0sinx= 0Question 6:Show that limx!0xsin(1x
) = 0 Answer:Note rst that for any real numbertwe have1sint1 so1sin(1x )1. Therefore, xxsin(1x )xand by squeeze theorem limx!0xsin1x = 0.Continuity and Dierentiability
Question 7:Which of the following functions areNOTeverywhere continuous: (A)f(x) =x24x+2 (B)f(x) = (x+ 3)4 (C)f(x) = 1066 (D)f(x) =mx+b (E) None of the above Answer:(A) Remember that, informally at least, acontinuousfunction is one in which there are nobreaks its curve. A continuous function can be drawn without lifting your pencil from the paper. More
formally, a functionf(x) iscontinuousat the pointx=aif and only if:1.f(x) is dened at the pointx=a,
2. the limit lim
x!af(x) exists,3. lim
x!af(x) =f(a) The functionf(x) =x24x+2is not everywhere continuous because the function is not dened at the point x=2. It is worth noting that limx!2f(x) does in fact exist!The existence of a limit at a point does not guarantee that the function is continuous at that point! 2 Question 8:Which of the following functions are continuous: (A)f(x) =jxj (B)f(x) =3x <4 12 x+ 3x4 (C)f(x) =1x (D)f(x) =lnx x <0 0x= 0 (E) None of the above Answer:(A) The absolute value functionf(x) =jxjis dened as: f(x) =x x0 x x <0 Does this function satisfy the requirements for continuity? Yes! The critical point to check isx= 0. Note that the function is dened atx= 0; the limx!0f(x) exists; and that limx!0f(x) = 0 =f(0). Question 9:Which of the following functions areNOTdierentiable: (A)f(x) =jxj (B)f(x) = (x+ 3)4 (C)f(x) = 1066 (D)f(x) =mx+b (E) None of the above Answer:(A) Remember that continuity is anecessarycondition for dierentiability (i.e., every dier- entiable function is continuous), but continuity is not asucientcondition to ensure dierentiability(i.e., not every continuous function is dierentiable). Case in point isf(x) =jxj. This function is in
fact continuous (see previous question). It is not however dierentiable at the pointx= 0. Why? The pointx= 0 is a cusp (or kink). There are an innite number of lines that could be tangent to the functionf(x) =jxjat the pointx= 0, and thus the derivative off(x) would have an innite number of possible values.Question 10:Is function
f(x) =0 :x= 0 xsin(1=x) :x6= 0 continuous at point 0?Answer:Note thatfis continuous at a pointaif
lim x!af(x) =f(limx!ax):In this case, we takea= 0 and
lim x!0f(x) = limx!0xsin(1=x) = 0 by question 6. Moreover, f(limx!0x) =f(0) = 0 thus,fis continuous at 0. 3Derivatives
Question 11:Find the derivative of the following function: f(x) = 1963 (A) +1 (B) 1963 (C)1 (D) 0 (E) None of the above Answer:(D) The derivative of a constant function is always zero. Question 12:Find the derivative of the following function: f(x) =x2+ 6x+ 9 (A)f0(x) = 2x+ 6 + 9 (B)f0(x) =x2+ 6 (C)f0(x) = 2x+ 6 (D)f0(x) = 2x (E) None of the aboveAnswer:(C) Remember that 1) the derivative of a sum of functions is simply the sum of the derivatives
of each of the functions, and 2) the power rule for derivatives says that iff(x) =kxn, thenf0(x) = nkx n1. Thusf0(x) = 2x21+ 6x11+ 0 = 2x+ 6. Question 13:Find the derivative of the following function: f(x) =x12 (A)f0(x) =12 px (B)f0(x) =1px (C)f0(x) =12 px (D)f0(x) =px (E) None of the above Answer:(C) Remember that the power rule for derivatives works with fractional exponents as well!