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Solve dis.
1. Electric vector E of an electromagnetic wave in a free space is given by Ex=Ez=0, and Ey=Acosw(t- z/c). By using the maxwell's equation for a free space, determine the expression for the components of H.

gbengarock: the guru in d house
differentiate y=x*x*x with respect to x
note: * = raised to power
fenks

i revere to u all to d maths general in dis house. i've missed so much here but all is well sha. i will nid to seat down nd review all wat i av missed . i still remain my humble name as rhydex247
soln.

let y=x^x^x.
which implies y=x^(x^x)
recall dat y=x^x
log bth sides we av
lny=lnx^x.... lny=xlnx....
d/dy(lny)=lnx+1
1/y(dy/dx)=(lnx+1)
dy/dx=y(lnx+1) buh y=x^x
dy/dx=x^x(lnx+1).
now dat we knw dy/dx of y=x^x we
can easily get y=x^(x^x)
log both sides we av
lny=lnx^(x^x)
lny=x^xlnx
d/dy(lny)=vdu/dx+udv/dx where
my u=x nd v=x^x.
dy/dx(1/y)=x^x(1)+x(x^x(lnx+1))
dy/dx=y[x^x+x(x^x( lnx+1))]
recall dat y=x^x^x
finally we have
dy/dx=x^x^x[x^x+x(x^x(lnx+1))]

[quote author=gowaga68]Greetings too all the guru's in the house.
Here are my questions for you to help me out (though got the answers and working i still need to know how its done).
1. In the set of all real numbers solve the equation
sqr root3X+10 - sqr root X+4=2 and find the conditions of solvability
SOLUTION.
√3x+10=2+√x+4
square both sides
3x+10=4+4√x+4+ x+4
2x+10=8+4√x+4
2x+2=4√x+4
squaring both sides yield
4x^2-8x-60=0
divide through by 4
x^2-2x-15=0
therefore x=-3 or x=5. lobatan

happy new year to u all my oga at the top.

Humphrey77: WHAT ARE ALGEBRAIC NUMBERS? :*@HAPPY

Algebraic numbers is a number that is a root of a non zero polynomial in one variable with rational coefficients or equivalent by clearing denominators with integer coefficients.

Humphrey77: GREAT! WHAT ARE MONIC EQUATION?

Do u mean harmonic equation?

lufemos: Wat r d criteria needed in statin d properties of a differential equation, ie how do I identify d order, degree, linearity and homogenity of a DE

hmmm.
for the order of differential eqn: is d order of d highest differential coefficient (derivative) contained in d eqn. e.g. d^2(y)/dx^2+2dy/dx+ydx=sinx is an example of 2nd order.

for the degree of differential eqn: is d power to which the highest order differential coefficient is raised when d eqn is rationalised i.e fractional power is removed. e.g (dy/dx)^3+y^2=x.
remaining soln loading by A.K.A SERIES.

Humphrey77: state and prove LAGRANGE THEOREM ( ABSTRACT ALGEBRA)

hmmmm.
dis is my best course
here the soln goes.
langrange theorem states that for any finite group G, the order number of element of every sub group H of G divides the order of G. The theorem is named after joseph louis langrange.

now the prove.
this can be shown using the concepts of left cosets of H in G. the left cosets are the equivalence classes of a certain equivalence relation on G and therefore form a partition of G. Specifically, x and y in G are related if and only if there exists h in H such that x=yh . If we can show dat all cosets of H av d same number of elements, then each coset of H has precisely /H/ elements. We are den done since the order of H times the number of cosets is equal to the no of elements in G, thereby proving that the order of H divides the order of G. Now, if aH and bH are two left cosets of H. We can define a map. f:aH--->bH. by setting f(x)=ba^-1x. This map is bijective because its inverse is given by f^-1 y=ab^-1(y).

2nioshine: Thanks for the approach
but note that the first principle required for Q1(not the shifting principle as used)*although a slight flaw in your ans.
And Q2 implies the laplace transform of the derivative i.e Lf'{}.( not so conventional) nd thats d reason i choose the notation f'L{ } ...nice try though...
Sti awaiting replies

hmmm. u mean d/dt[L(e^4tcos3t)]. if that is d case solution pending.

2nioshine: Topic:LAPLACE TRANSFORMATION

Q1
find L{tcosat},where a=constant. from first principle
Q2
find f 'L{e^(4t)cos3t}

Nb:f ' means the derivative of the function

solution to no 1.
L(tcosat) using the shift theorem.
well known that L(cosat)= s/s^2+a^2 where s>0.
now d L(tcosat)= -d/ds(s/s^2+a^2).
Using d quotient rule we av s^2-a^2/(s^2+a^2)^2.

solution to no 2.
L(e^4tcos3t). well known that L(Cos3t)=s/s^2+9.
hence L(e^4tcos3t)= s-4/(s-4)^2+9= s-4/s^2-8s+25.
now d f'L(e^4tcos3t) = f'[ (s-4)/(s^2-8s+25)].
using d quotient rule
we ave s^2-8s+7/(s^2-8s+25)^2.

smurfy: Why is everybody scared of tackling my simple permutation question?

I'll post it forever until... Yeah, you got it.

DO U TINK UR QUESTION ARE HARD.
MAKE I REACH HOUSE U'LL GET THE SOLN DIS NITE.

@ MR BENBUKS.
MY CARD NA. WIT D PREVIOUS CARD O MAKING 200 NAIRA. I DEY PARA O.
SMURFY NA SHORTCUT WE TAKE SOLVE D QUESTION.

benbuks: Tym up.

where my recharge card?

benbuks: . . . No sir, hint[Q1 put x=tan@)
Q2 put x =cos@]

for d no 2. it doesnt meam we shud use d same method. moreover u did not state dat in d beginning of ur questn.

smurfy: Wrong. Mine is correct. I want etisalat and not mtn o.

av modify my ans.

[quote author=benbuks]1) d/dx [arccos (1-x^2)/(1+x^2) ]

i can only provide d ans cos my ba3 is drained.
the ans is 2sqrt[x^2/(x^2+1)^2]/x.

benbuks: 1) d/dx [arccos (1-x^2)/(1+x^2) ]

2) d/dx[arccos(4x^3 - 3x)]

burst your brains with the above enigma... Should i put a prize tag?

let y= arccos 4x^3-3x
also let u=4x^3-3x...du/dx=12x^2-3
y=arccosu
dy/du=1/ sqrt(1-u^2)
dy/dx=12x^2-3/sqrt(1-(4x^3-3x)^2)

benbuks: 1) d/dx [arccos (1-x^2)/(1+x^2) ]

2) d/dx[arccos(4x^3 - 3x)]

burst your brains with the above enigma... Should i put a prize tag?

yes put a price tag

benbuks: . . .i said b4 27mins (<1pm) nt </=1pm,.. Bt any hw sha ..2 avoid problems...meet me 4 2go 4 ur prize.

lol. 100 naira recharge card. i go take am call my aristo gf.

benbuks: ^^Time up...ma kip d card 4 anoda question.

d tym is not up b4 pastin my soln.
my post no is 3768 while urs is 3769. guy no do partia o

benbuks: X^4 + y^4 =82 .....(a)
x+y = 4 .....(b)
solve

(N100 mtn) first to post correct solution and final answer b4
27mins
show full workings , else nothing for you

from eqn ii x=4-y. put x=4-y in eqn i
(4-y)^4+y^4=82.
by expansn we av
y^4-16y^3+96y^2-256y+256+y^4=82.
2y^4-16y^3+96y^2-256y+174=0
divide thru by 2
y^4-8y^3+48y^2-128y+87=0
(y-1)(y-3)(y^2-4y+29)=0
y=1 or y=3
recall dat x=4-y .
x=1 or x=3. for d y^2-4y+29=0 will lead to complex ans.

hardedeji1: this particular line is whr u got it wrong...


dy/dx(1/y)=x^x(1)+x(x^x(lnx+1))

differential of x^xlnx should b

(x^x)/x + lnx[(x^x)(lnx + 1)]

recheck ur workings, note dat d/dx(lnx) is 1/x and not 1 as u wrote.

thanks for dat. d error is frm typing

[quote author=Calculusf(x)].i respect you bro,but here is my approach...y=x^x^x...y=x^(x^2).try this with calculator and substitute values like 2 and 3 and check if it's correct...for 2...x^x^x=16 and x^(x^2)=16 try that of 3 also...and y=x^x^x means...y={[(x)^x]^x} and from indices u=(x^a)^b=x^ab...then y=x^(x.x)=x^(x^2)...so from the question y=x^x^x...taken natural log of both sides...lny=ln.x^x^x...lny=xlnx^x...(lnx^x=xlnx)then lny=x.xlnx...lny=x^2lnx...1/y.dy/dx=x+2xlnx...dy/dx=x^x^x{x+2xlnx}[/quote]hmmmm.
here is my skelewu approach.
y=x^x^x.
which implies y=x^(x^x)
recall dat y=x^x
log bth sides we av
lny=lnx^x.... lny=xlnx....
d/dy(lny)=lnx+1
1/y(dy/dx)=(lnx+1)
dy/dx=y(lnx+1) buh y=x^x
dy/dx=x^x(lnx+1).
now dat we knw dy/dx of y=x^x we can easily get y=x^(x^x)
log both sides we av
lny=lnx^(x^x)
lny=x^xlnx
d/dy(lny)=vdu/dx+udv/dx where my u=x nd v=x^x.
dy/dx(1/y)=x^x(1)+x(x^x(lnx+1))
dy/dx=y[x^x+x(x^x( lnx+1))]
recall dat y=x^x^x
finally we have
dy/dx=x^x^x[x^x+x(x^x(lnx+1))] shikena.

smurfy: Factorise completely 20x^4 + 154x^3 + 40x^2 - 418x - 84.

the aswer is
2(x+2)(x+7)(2x-3)(5x+1). shikena.

smurfy: Find dy/dx if y = (arcsinx)^cos^2(x).

HMMM. HERE IS THE ANS I WILL POST THE WORKINGS TOMORO.
(Sin^-1x)^(cos^2x)[ {cos^2x/ √1-x^2sin^-1x} - (2sinxcosx(lnsin^-1x))]… na pingin pingin tins on point.

kwakayekaa: Let Q>0 be the set of positive rational numbers. Let f : Q>0 ! R be a function satisfying
the following three conditions:
(i) for all x; y 2 Q>0, we have f(x)f(y)  f(xy);
(ii) for all x; y 2 Q>0, we have f(x + y)  f(x) + f(y);
(iii) there exists a rational number a > 1 such that f(a) = a.
Prove that f(x) = x for all x 2 Q>0.

hmmmmm. real analysis again.
make i run carry my olubunmo or bartle G. Sherbert.

good mawnin to u all.

Compute the derivative of..
**] sqrt[ (x^2 +1)^2 +sqrt (1 + (x^2 +1)^2 ) ]
nb:"sqrt." Means square root[/quote]SOLUTION BY A.K.A SERIES.
HERE IT GOES.
USING THE CHAIN RULE,
d/dx(sqrt[(x^2+1)^2+sqrt(1+(x^2+1)^2)]=dsqrt(u)/du*du/dx , where u=(x^2+1)^2+sqrt(1+(x^2+1)^2)
and d/du(sqrt(u)= 1/2sqrt(u).
d/dx[(x^2+1)^2+sqrt(1+(x^2+1)^2)]/2sqrt[(x^2+1)^2+sqrt(1+(x^2+1)^2)].
differentiatin the sum term by term
d/dx[(x^2+1)^2]+d/dx[sqrt(1+(x^2+1)^2)]/2sqrt[(x^2+1)^2+sqrt(1+(x^2+1)^2)].
Using the chain rule, d/dx[(x^2+1)^2]= du^2/du*du/dx, where u=x^2+1 and d/du(u^2)=2u.
d/dx[sqrt(1+(x^2+1)^2)]+4x^3+4x/2sqrt[(x^2+1)^2+sqrt(1+(x^2+1)^2)].
Using chain rule d/dx[sqrt(1+(x^2+1)^2)] d/du(sqrt(u))*du/dx, where u=1+(x^2+1)^2 and d/du(sqrt(u))=1/2sqrt(u).
4x^3+4x+[d/dx(1+(x^2+1)^2)/2sqrt(1+(x^2+1)^2)]/2sqrt[(x^2+1)^2+sqrt(1+(x^2+1)^2)].
4x^3+4x+[d/dx(1)+d/dx(x^2+1)^2)]/2sqrt(1+(x^2+1)^2)]/2sqrt[(x^2+1)^2+sqrt(1+(x^2+1)^2)].
4x^3+4x+[2(2x)(x^2+1)/2sqrt(1+(x^2+1)^2)]/2sqrt[(x^2+1)^2+sqrt(1+(x^2+1)^2)].
4x^3+4x+[2x(x^2+1)/sqrt(1+(x^2+1)^2)]/2sqrt[(x^2+1)^2+sqrt(1+(x^2+1)^2)].
that's all. All is well.

[quote author=Humphrey77]if 3+5 qual 8 prove that 1+1 equal 2


solution.
this is real analysis.
here it goes.
well known that 1.a=a from the axiom ii.
from 1+1=2
1.a+1.a=1(a+a)=2a. hence 1+1=2 Q.E.D

the soln to dis questn long o

benbuks: ...bravo,...try d 2nd 1.....kinda long sha..

re write the second question bro.

@ benbuks.
ur question d/dx( sqrt[1+sqrt(1+sqrt(x)].
here is my solution.
d/dx( sqrt[1+sqrt(1+sqrt(x)]= d/du[rut(u)]*du/dx... where u=sqrt[1+sqrt(x)]+1... d/du(sqrt(u)= 1/2sqrt(u).
d/dx(1+sqrt[1+sqrt(x)]/2sqrt[1+sqrt(1+sqrt(x)]
differentiate the sum term by term.
d/dx(1)+d/dx(sqrt[1+sqrt(x)]/2sqrt[1+sqrt(1+sqrt(x)].
the derivative of 1 is zero we av
d/dx(sqrt[1+sqrt(x)]/2sqrt[1+sqrt(1+sqrt(x)].
Using the chain rule, d/dx(sqrt[1+sqrt(x)])=d/du(sqrt(u)*du/dx, where u=sqrt(x)+1 and d/du(sqrt(u))=1/2sqrt(u)
d/dx(1+sqrt(x))/2sqrt[1+sqrt(x)]/2sqrt[1+sqrt(1+sqrt(x)].
simplifying the expression gives
d/dx(1)+d/dx(sqrt(x))/4sqrt[1+sqrt(1+sqrt(x)][sqrt[1+sqrt(x)].
which gives
1/2sqrt(x)/4sqrt[1+sqrt(1+sqrt(x)][sqrt[1+sqrt(x)].
simplifying the expression gives
1/8sqrt[1+sqrt(1+sqrt(x)][sqrt[1+sqrt(x)][sqrt(x)].
thats all. all is well