DIFFERENTIAL EQUATIONS
(ii) A differential equation involving derivatives of the dependent variable with x. ∫. ⇒ logy = logx + logc ⇒ y = cx. Example 3 Given that.
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CONTINUITY AND DIFFERENTIABILITY
(iii) Every differentiable function is continuous but the converse is not true The derivative of logx. w.r.t.
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6.2 Properties of Logarithms
4. log 3. √. 100x2 yz5. 5. log117(x2 − 4). Solution. 1. To expand log2. (8 x) we use the Quotient Rule identifying u = 8 and w = x and simplify.
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4-Partial Derivatives and their Applications.pdf
log (. 3 ) u x y z xyz x x. ∂. ∂. +. + −. ∂. ∂. (i.e. partial derivative of u with respect to x keeping y and z– constant). 3. 3. 3. 3. 3. 3.
Partial Derivatives and their Applications
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Week 3 Quiz: Differential Calculus: The Derivative and Rules of
Week 3 Quiz: Differential Calculus: The Derivative and Rules of Differentiation. SGPE Summer School 2016. Limits. Question 1: Find limx→3f(x): f(x) =.
week answers
Week #3 - Exponential Functions and Logarithms; The Derivative
QUIZ PREPARATION PROBLEMS. 6. For the function f(x) = log(x) estimate f′(1). From the graph of log(x)
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Answers to Exercises
Product 10 - 15 (sin x 3)(1 / 2x 2)4x - (log 2X2)(COS x3)3x2. 59. . 3 2 ... find the n-th derivative we just divide n by 4
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11- Partial Differential Equations.pdf
Taking partial derivative of equation (3) with respect to x (i) u = x2 – y2
Partial Differential Equations
University of Plymouth
25 mai 2005 3. Higher Order Partial Derivatives. 4. Quiz on Partial Derivatives ... Example 3 Find. ∂z. ∂x for each of the following functions.
PlymouthUniversity MathsandStats partial differentiation
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 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- log base x 3 derivative
- log x 3 differentiation
- derivative of log 3 x^2
- nth derivative of log x^3