Differentiating logarithm and exponential functions
differentiate ln x from first principles. • differentiate ex. Contents. 1. Introduction. 2. 2. Differentiation of a function f(x).
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EXERCISES IN MATHEMATICS Series F No. 2: Answers First
1. Differentiate from first principles y = x2 − 4x. Answer. We have y + δy = (x + δx)2 − 4(x
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log y. = = 0720 log (cos x x) (0. Lt sin x lim. →0 log y = 0. COS X Differentiation from First Principle (AB-Initio Method). Let f(x) is a function ...
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1 Theory of convex functions
1 mar. 2016 Let's first recall the definition of a convex function. ... In words this means that if we take any two points x
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CONTINUITY AND DIFFERENTIABILITY
The derivative of logx. w.r.t. x is. 1 x. ; i.e.. 1. (log ) d x dx x. = . 5.1.12 Logarithmic differentiation is a powerful technique to differentiate
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Recapitulation of Mathematics
is known as the first principle of differential calculus. 1.2 Differential Coefficient of a Function at a Point. The value of the derivative of f(x)
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One Variable Calculus with SageMath
13 juil. 2019 Find the derivative of h(x) = log(x) + x5 + sin(x) using the first principle. sage: ax=var('a
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Week 3 Quiz: Differential Calculus: The Derivative and Rules of
is important to note the this function is undefined at x = 3. Answer: Note first that for any real number t we have −1 ≤ sint ≤ 1 so −1 ≤ sin(1 x. ) ...
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Course Contents: Topic and Contents Hours Marks : 15SC02M
Derivatives of functions ofx sin x
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Differentiating logarithm and exponential functions
differentiate ln x from first principles. • differentiate ex. Contents. 1. Introduction. 2. 2. Differentiation of a function f(x).
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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
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
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