₦airaland Forum

agentofchange1:
Evaluate

|5 1 0 0 0|
|1 4 0 1 2|
|3 2 1 4 0|
|0 1 2 1 0|
|1 2 0 1 2|

naturalwaves:
Is this a 5*5 matrix and do you want us to find the determinant?

agentofchange1:
yes. sir

agentofchange1:
Evaluate

|5 1 0 0 0|
|1 4 0 1 2|
|3 2 1 4 0|
|0 1 2 1 0|
|1 2 0 1 2|

naturalwaves:
Well, I got. "-140" as my answer. If that is wrong, then, I must have made a mistake in the calculation. See my solution below,

Nastydroid:
if I should use 3x3 method will I arrive at the same ans

naturalwaves:
Well, I got. "-140" as my answer. If that is wrong, then, I must have made a mistake in the calculation. See my solution below,

naturalwaves:
Well, I got. "-140" as my answer. If that is wrong, then, I must have made a mistake in the calculation. See my solution below,

agentofchange1:
seems or right ....leme verify something . sha ..




BTW:still expecting my airtime ooo.

Drniyi4u:
ţђųmbş ųp!

yemstok:
Pls guys kindly help out with these
'Derivative of logarithm functions'

1. y=a^cosx

2. y=x^x

3. y=x^xsquare

4. y=e^1+x/1-x

5. y=2^inx

agentofchange1:
**Solution **

1.)

put t=cosx ; dt/dx =-sinx
=>y=a^t dy/dt =a^t *lna
dy/dx =dt/dx* dy/dt

=>dy/dx =-a^t *sinx * lna

thus , dy/dx =-a^cosx sinx *lna


2.)

taking Napierian logarithm of both sides , we have

lny=xlnx

put F(x,y)=lny-xlnx

f_x=-lnx -1

f_y=1/y

now dy/dx=-f_x /f_y

=>dy/dx =-(-lnx-1)/(1/y)
dy/dx= y(lnx+1)

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

3.) similarly

take natural logarithm of both sides

lny = x² lnx

y'/y = 2xlnx + x

y'=dy/dx = y(2xlnx +x)

replace 'y'


4.) y= e^{[ 1+ x/(1-x)]}
put
u =[1 + x/(1-x)] =1/(1-x) ; du/dx =(1-x)^-2

now y' = dy/dx = u'e^u (where u '= du/dx )

dy/dx =(1-x)^-2 * e ^{[1 + x/(1-x) ]}

………(replace u)

5.) y=2^lnx

put u = lnx ; du/dx = 1/x

=> y= 2^u dy/du = 2^u *ln2

now y' = 2^u*ln2 * 1/x

hence dy /dx = 2^lnxln2 /x



that's it man

yemstok:
Ure the best man! Gracias bro!

Arithmetic:
If tan²x = 1 + 2tan²y. Show cos2x + sin²y = 0. Thanks…

badmus45:
1. The equation of motion of a forced oscillation is d2y/dt+2dy/dt
+5y=4sin5t solve the equation given that t=0 ,y=0 and dy/dt=0.
(2)determine the equation of the curve which satisfies the equation
xydy/dx=x^2-1 which passes. thru point (1,2)........abeg

Laplacian:
recall, sec²y=1+tan²y.
From the LHS,
sec²x-1=1+2(sec²y-1) or
sec²x=2sec²y or
cos²y=2cos²x or
cos²y=(2cos²x-1)+1, does the expression in the bracket look familiar? *****





That's cos2x.
So, cos²y=cos2x+1
or
cos²y-1=cos2x
can u finish it from here? Hop it helps!

Arithmetic:
1). $[1 / {3^x + e^-x}]dx.
2). $[log(1-x)/e^-2x]dx. Thanks...