Solve the equation cos^2(x) +cos^2 (2x) + cos^2 (3x) =1

@citizen! tha[nk usir n God bless u . quote author= d citizen]Q2(a),

Y= 2x^3 -5x^2 + 4x -1

dy/dx=6x^2-10x + 4

At the stationary point, dy/dx =0

6x^2- 10x + 4= 0

6x^2 -6x -4x + 4=0

6x (x-1) -4(x-1)=0
(6x-4)(x-1)=0
x=1 or x=4/6=2/3

But, d^2y/dx^2 = 12x-10

Wen x=1, d^2y/dx^2= 12(1)-10=2 (minimum becos d^2y/dx^2 >0

Ymin= 2(1)^3 -5(1)^2 +4(1)-1=0

Ymax=2(2/3)^3 -5(2/3)^2 + 4(2/3)-1
. Calculate it

y=x^4+2x^3- 3x^2 -4x 6.

dy/dx=4x^3+6x^2-6x-4.

d^2y/dx^2= 12x^2 +12x-6.

d^3y/dx3= 24x- 12

Since d^3y/dx^3=0,

24x-12=0
X=-12/24=-1/2. right on board, the math doctor

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benbuks: Solve the equation cos^2(x) +cos^2 (2x) + cos^2 (3x) =1

using multiple angle formulae, we can restructure the question to become (cos²x)+(2cos²x-1)²+(4cos³x-3cosx)²....... replacing cos²x= n and expanding the bracket we have 4n³-20n²+6n=0.......... factorizing and dividing through by 2, we get a quadratic equation....... 2n²-10n+3= 0, using formula method, the values for n are 0.5 and 4.5, we discard the second value since the maximum value of cosx is1.......... finally x= cos^-1(sqrt.0.5)........ therefore x= 45^o or 315^o

honey: using multiple angle formulae, we can restructure the question to become (cos²x)+(2cos²x-1)²+(4cos³x-3cosx)²....... replacing cos²x= n and expanding the bracket we have 4n³-20n²+6n=0.......... factorizing and dividing through by 2, we get a quadratic equation....... 2n²-10n+3= 0, using formula method, the values for n are 0.5 and 4.5, we discard the second value since the maximum value of cosx is1.......... finally x= cos^-1(sqrt.0.5)........ therefore x= 45^o or 315^o

..nice 1

Try

QUESTION F(X)=3/X^3. Using ist principle method
SOLUTION
Let y=3/x^3
i will let my change in x =h and change in y =k
y+k=x+h
y+k=3/(x+h)^3
y+k=3/x^3+3x^2h+3xk^2+k^3
subtract y from both sides
y+k-y=3/x^3+3x^2h+3xh^2+h^3 -y
but y=3/x^3
k=3/x^3+3x^2h+3xh^2+h^3 -3/x^3
taking the l.c.m
k=3x^3 - 3(x^3+3x^2h+3xh^2+h^3) /x^3(x^3+3x^2h+3xh^2+h^3)
k=3x^3 - 3x^3-9x^2h-9xh^2-3h^3 /x^3(x^3+3x^2h+3xh^2+h^3)
k=-9x^2h-9xh^2-3k^3/x^3(x^3+3x^2h+3xh^2+h^3)
divide through by change in x=h
k/h= -9x^2h-9xh^2-3h^3/x^3(x^3+3x^2h+3xh^2+h^3)*1/h
k/h= -9x^2-9xh-3h^2/x^3(x^3+3x^2h+3xh^2+h^3)
k/h=dy/dx=lim h---->0[-9x^2-9xh-3h^2/x^3(x^3+3x^2h+3xh^2+h^3)]
dy/dx=-9x^2/x^3(x^3)
dy/dx=-9x^2/x^6=-9/x^4
therefore dy/dx=-9x^-4. good luck


QUESTION Lim--->x=pi(sin(picosx)/(x-pi)^2)
solution
we all know that pi=180
putx=pi=180
we obtain 0/0. which is indeterminate.
using l'hopitals rule which says
limx--->a (f'(x)/g'(x))
when you differentiate sin(picosx) w.r.t.
i.e f'(x)=cos(picosx)(-pisinx)
g'(x)= 2(x-pi)
limx--->pi[cos(picosx)(-pisinx)/2(x-pi)]
when we put x=pi we obtain 0/0 which is indeterminate
we will differentiate again untill we obtain a real value
f''(x)=cos(picosx)(-picosx)+(-pisinx)(sin(picosx))(pisinx)
g''(x)= 2
limx---->pi[(cos(picosx)(-picosx)+(-pisinx)(sin(picosx)(pisinx)/2]
putx=pi=180
we obtain [(-1)(180)+0/2]
which implies -90 or -pi/2.

Integrate(e^tanx)dx

2^x + 3^y=z^2 solve..

If you think u r a math general chk out pass questions in british mathematical olympaid (BMO)

Solve Log 64 base 16 + Log 81 base 1/3 + Log 32 base 1/2

X+y =5
x^x + y^y=31.......find x and y....

Fetus: X+y =5
x^x + y^y=31.......find x and y....

..(2,3)

Olarewajub: Solve Log 64 base 16 + Log 81 base 1/3 + Log 32 base 1/2

...since dia of diff. Base so evaluate individually then carry out the operation..log 64 base 16 =log 8 base 4=3/2
log81 base 1/3 =-4
log 32 base 1/2=-5 hence 3/2 -9/1 =-15/2

Help solve dis o-the numerator of a fraction is 5 less than its denominator.if 6 is added to the numerator and 4 to the denominator,the fraction is doubled.what is the fraction?Pls, help me out with this:

busuyem: Help solve dis o-the numerator of a fraction is 5 less than its denominator.if 6 is added to the numerator and 4 to the denominator,the fraction is doubled.what is the fraction?Pls, help me out with this:

3/8

I don't think there's an analytical solution. for integral e^tanxdx

If you try it with Wolfram's online integrator, it gives the result in terms of the Exponential Integral Ei function, which I believe is transcendental.

integral e^tanxdx
solution
using wolfram mathematical online integrator
-1/2ie^-id[(e^2iEi(tanx-i)-Ei(tanx+i)].

prove that 1=2
solution
1 can never be equal to 2. hence the answer is FALSE.
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[color=#990000][/color]

benbuks: ..(2,3)

...tanx man..bt can u show d workings plz..

x + y = 5.......(1)
x^x + y^y =31.......(2)

from eqn(1), x=5-y, hence we substitute this value for x in eqn(2).
(5-y)^(5-y) + y^y = 31.......(3)
now, we have reduced the problem to what value of y that must be inserted in eqn (3) to obtain 31.

a thorough look at eqns (1) & (2) shows that the values of x & y must be positive integers within the range 0<x<5 and 0<y<5
clearly, y=3
i.e (5-3)^(5-3) + 3^3
2^2 + 27
4+27=31
now that we have established that y=3, we shall substitute the value of y in eqn(1) to get x
recall that x+y=5
therefore; x+3=5 or x=5-3=2
hence, x=2 when y=3. therefore we av (2,3)

Differentiate y= x^x^x^x^x to infinte with respect to x . I pose dis question to the math general

Ortarico: ^^ @Doubledx
Plz i came across this challenge in one of the threads:

1. A GP of a positive terms and an AP have same first term. The sum of their first term is 3, the sum of their second term is 3/2, and the sum of their third term is 6. Find the sum of their fifth term.

2. A trader bought 30 articles, he sold 20 of them at a gain of 16% and sold the remainder at a loss of 4%. Find the percentage profit/loss on the articles.

3. Find the value of x if (x+3)^3 + (x+4)^3=(2x+7)^3

Guyz do something. . .

questions 1&2 are straight forward, for question 3, set x=y-7/2, generally, always take d mean of d first two brackets, equate to zero, then u solve 2 get d constant 2 b added 2 d new variable y.

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Derivative:

d\/dx(x^(x^(x^(x^x)))) = x^(x^(x^(x^x))) (x^(x^(x^x)-1)+x^(x^(x^x)) log(x) (x^(x^x-1)+x^(x^x) log(x) (x^(x-1)+x^x log(x) (log(x)+1)))

dx(x^(x^(x^(x^x))))
Express x^(x^(x^(x^x))) as a power of e: x^(x^(x^(x^x))) = e^(log(x^(x^(x^(x^x))))) = e^(x^(x^(x^x)) log(x))
= d\/dx(e^(x^(x^(x^x)) log(x)))
Using the chain rule, d\/dx(e^(x^(x^(x^x)) log(x))) = de^u/du* du\dx, where u = x^(x^(x^x)) log(x) and d\du(e^u) = e^u = d\dx(x^(x^(x^x)) log(x))) e^(x^(x^(x^x)) log(x))
Express e^(x^(x^(x^x)) log(x)) as a power of x: e^(x^(x^(x^x)) log(x)) = e^(log(x^(x^(x^(x^x))))) = x^(x^(x^(x^x)))
= x^(x^(x^(x^x))) d\/dx(x^(x^(x^x)) log(x))
Use the product rule, d/dx(u,v) = v du/dx+udv)/dx, where u = x^(x^(x^x)) and v = log(x)
= log(x) d\/dx(x^(x^(x^x)))+x^(x^(x^x)) d\/dx(log(x)) x^(x^(x^(x^x)))
Express x^(x^(x^x)) as a power of e: x^(x^(x^x)) = e^(log(x^(x^(x^x)))) = e^(x^(x^x) log(x))
= x^(x^(x^(x^x))) (x^(x^(x^x)) (d/dx(log(x)))+d/dx(e^(x^(x^x) log(x))) log(x))
Using the chain rule, d/dx(e^(x^(x^x) log(x))) =de^u/du* du\dx,where u = x^(x^x) log(x) and d/du(e^u) = e^u
= x^(x^(x^(x^x))) (x^(x^(x^x)) (d/dx(log(x)))+d/dx(x^(x^x) log(x)) e^(x^(x^x) log(x)) log(x))
Express e^(x^(x^x) log(x)) as a power of x: e^(x^(x^x) log(x)) = e^(log(x^(x^(x^x)))) = x^(x^(x^x))
= x^(x^(x^(x^x))) (x^(x^(x^x)) d/dx(x^(x^x) log(x)) log(x)+x^(x^(x^x)) (d/dx(log(x))))
Use the product rule, d/dx(u, v) = v ( du)/( dx)+u ( dv)/( dx), where u = x^(x^x) and v = log(x)
= x^(x^(x^(x^x))) (x^(x^(x^x)) (d\/dx(log(x)))+log(x) d\/dx(x^(x^x))+x^(x^x) d\/dx(log(x)) x^(x^(x^x)) log(x))
Express x^(x^x) as a power of e: x^(x^x) = e^(log(x^(x^x))) = e^(x^x log(x))
= x^(x^(x^(x^x))) (x^(x^(x^x)) (d/dx(log(x)))+x^(x^(x^x)) log(x) (x^(x^x) (d/dx(log(x)))+d/dx(e^(x^x log(x))) log(x)))
The derivative of log(x) is 1/x
= x^(x^(x^(x^x))) (x^(x^(x^x)) log(x) (x^(x^x) (d/dx(log(x)))+log(x) (d/dx(e^(x^x log(x)))))+1/x x^(x^(x^x)))
Simplify the expression gives
= x^(x^(x^(x^x))) (x^(-1+x^(x^x))+x^(x^(x^x)) log(x) (x^(x^x) (d/dx(log(x)))+log(x) (d/dx(e^(x^x log(x))))))\nThe derivative of log(x) is 1\/x:\n = x^(x^(x^(x^x))) (x^(-1+x^(x^x))+x^(x^(x^x)) log(x) (log(x) (d\/dx(e^(x^x log(x))))+1\/x x^(x^x)))\nSimplify the expression:\n = x^(x^(x^(x^x))) (x^(-1+x^(x^x))+x^(x^(x^x)) log(x) (x^(-1+x^x)+log(x) (d\/dx(e^(x^x log(x))))))\nUsing the chain rule, d\/dx(e^(x^x log(x))) = ( de^u)\/( du) ( du)\/( dx), where u = x^x log(x) and ( d)\/( du)(e^u) = e^u:\n = x^(x^(x^(x^x))) (x^(-1+x^(x^x))+x^(x^(x^x)) log(x) (x^(-1+x^x)+d\/dx(x^x log(x)) e^(x^x log(x)) log(x)))\nExpress e^(x^x log(x)) as a power of x: e^(x^x log(x)) = e^(log(x^(x^x))) = x^(x^x):\n = x^(x^(x^(x^x))) (x^(-1+x^(x^x))+x^(x^(x^x)) log(x) (x^(-1+x^x)+x^(x^x) d\/dx(x^x log(x)) log(x)))\nUse the product rule, d\/dx(u v) = v ( du)\/( dx)+u ( dv)\/( dx), where u = x^x and v = log(x):\n = x^(x^(x^(x^x))) (x^(-1+x^(x^x))+x^(x^(x^x)) log(x) (x^(-1+x^x)+log(x) d\/dx(x^x)+x^x d\/dx(log(x)) x^(x^x) log(x)))\nExpress x^x as a power of e: x^x = e^(log(x^x)) = e^(x log(x)):\n = x^(x^(x^(x^x))) (x^(-1+x^(x^x))+x^(x^(x^x)) log(x) (x^(-1+x^x)+x^(x^x) log(x) (x^x (d\/dx(log(x)))+d\/dx(e^(x log(x))) log(x))))\nUsing the chain rule, d\/dx(e^(x log(x))) = ( de^u)\/( du) ( du)\/( dx), where u = x log(x) and ( d)\/( du)(e^u) = e^u:\n = x^(x^(x^(x^x))) (x^(-1+x^(x^x))+x^(x^(x^x)) log(x) (x^(-1+x^x)+x^(x^x) log(x) (x^x (d\/dx(log(x)))+d\/dx(x log(x)) e^(x log(x)) log(x))))\nExpress e^(x log(x)) as a power of x: e^(x log(x)) = e^(log(x^x)) = x^x:\n = x^(x^(x^(x^x))) (x^(-1+x^(x^x))+x^(x^(x^x)) log(x) (x^(-1+x^x)+x^(x^x) log(x) (x^x d\/dx(x log(x)) log(x)+x^x (d\/dx(log(x))))))\nUse the product rule, d\/dx(u v) = v ( du)\/( dx)+u ( dv)\/( dx), where u = x and v = log(x):\n = x^(x^(x^(x^x))) (x^(-1+x^(x^x))+x^(x^(x^x)) log(x) (x^(-1+x^x)+x^(x^x) log(x) (x^x (d\/dx(log(x)))+log(x) d\/dx(x)+x d\/dx(log(x)) x^x log(x))))\nThe derivative of x is 1:\n = x^(x^(x^(x^x))) (x^(-1+x^(x^x))+x^(x^(x^x)) log(x) (x^(-1+x^x)+x^(x^x) log(x) (x^x (d\/dx(log(x)))+x^x log(x) (x (d\/dx(log(x)))+1 log(x)))))\nThe derivative of log(x) is 1\/x:\n = x^(x^(x^(x^x))) (x^(-1+x^(x^x))+x^(x^(x^x)) log(x) (x^(-1+x^x)+x^(x^x) log(x) (x^x log(x) (log(x)+x (d\/dx(log(x))))+1\/x x^x)))\n
Simplify the expression= x^(x^(x^(x^x))) (x^(-1+x^(x^x))+x^(x^(x^x)) log(x) (x^(-1+x^x)+x^(x^x) log(x) (x^(-1+x)+x^x log(x) (log(x)+x (d\/dx(log(x)))))))\nThe derivative of log(x) is 1\/x:\n = x^(x^(x^(x^x))) (x^(-1+x^(x^x))+x^(x^(x^x)) log(x) (x^(-1+x^x)+x^(x^x) log(x) (x^(-1+x)+x^x log(x) (log(x)+1\/x x))))\n
Simplify the expression:
Answer: = x^(x^(x^(x^x))) (x^(-1+x^(x^x))+x^(x^(x^x)) log(x) (x^(-1+x^x)+x^(x^x) log(x) (x^(-1+x)+x^x log(x) (1+log(x)))))
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@Rhydex,

You are totally wrong, math major. You are not a math general yet. Keep trying your best.

The question still read differentiate y= x raise x raise x raise x raise x and so on with respect to x.

d/dx(x^x^x^x^x)=x^x^x^x^x+x^x^x[x^x^xlogx(x^xlogx(1/x+log^2x+logx)+1/x)+1/x].

Adokwu.Ondoma:


questions 1&2 are straight forward, for question 3, set x=y-7/2, generally, always take d mean of d first two brackets, equate to zero, then u solve 2 get d constant 2 b added 2 d new variable y.

hmmmm, No 1 question is from further maths project 1, i asked my teacher that can a question combine both A.P and G.P together and he said yes, unfortunately 4 he gave it to me alone as an assignment, if i could still remember my ans was 1024, can't recall sha

Differentiate y=x^x^x^x^ and so on with respect to x.

Analysis

y= x^x^x^x^x and so on

Thus, y=x^y

Introduce in to both sides:

In y= in x^y

Iny=yinx

Differentiate both sides:

i/y dy/dx = y * i/x inx * dy/dx (product rule)

Collect like terms:
dy/dx( 1/y -inx) = y/x
dy/dx =y/x/ (1/y - inx)
Simple, the math general.

d citizen:
Differentiate y=x^x^x^x^ and so on with respect to x.

Analysis

y= x^x^x^x^x and so on

Thus, y=x^y

Introduce in to both sides:

In y= in x^y

Iny=yinx

Differentiate both sides:

i/y dy/dx = y * i/x inx * dy/dx (product rule)

Collect like terms:
dy/dx( 1/y -inx) = y/x
dy/dx =y/x/ (1/y - inx)
Simple, the math general.

hmm......bro though am still a learner i dont totally agree wit u.
if y=2^2 u cant say y=2^y.
so dat place u said y=x^x^x^ u cant say y=x^y

@ d citizen
hint on how to solve y=x^x^x^x^x.
soln.
let me start 4rm y=x^x. Take log of both sides.
lny=xlnx. d/dx(lny)=d/dx(xlnx)
1/y(dy/dx)=lnx+1.
dy/dx=y(lnx+1). Recall that y=x^x.
dy/dx=x^x(lnx+1).
Similarly for y=x^(x^x).
Take log of both sides. We av
lny=lnx^(x^x). Which implies.
lny=x^xlnx. d/dx(lny)=d/dx(x^xlnx).
dy/dx(1/y)= d/dx(x^x)*lnx+d/dx(lnx)x^x.
now recall that d/dx(x^x)=x^x(lnx+1) and d/dx(lnx)=1/x.
dy/dx(1/y)=x^x(lnx+1)lnx+1/x*x^x.
dy/dx=y[x^x(lnx+1)lnx+x^-1+x]. Recall that y=x^x^x. Hence we av dy/dx=x^x^x[x^x(lnx+1)lnx+x^(x-1)].
use this method for y=x^(x^x^x) and y=x^(x^x^x^x). All is well.

Adokwu.Ondoma:


questions 1&2 are straight forward, for question 3, set x=y-7/2, generally, always take d mean of d first two brackets, equate to zero, then u solve 2 get d constant 2 b added 2 d new variable y.

Hmmmm. . .Although solved, thank you.

Boladearo: hmmmm, No 1 question is from further maths project 1, i asked my teacher that can a question combine both A.P and G.P together and he said yes, unfortunately 4 he gave it to me alone as an assignment, if i could still remember my ans was 1024, can't recall sha

It's 105.