Mathematics: Post your doubts here!

[whitecorp]

how do i differentiate ln(xy) with respect to x???

This is done implicitly. d/dx [ ln(xy)] = 1/(xy) * [ x (dy/dx) +y ]

Hope this helps. Peace.

 

[Alice123]

This is done implicitly. d/dx [ ln(xy)] = 1/(xy) * [ x (dy/dx) +y ]

Hope this helps. Peace.

thanks
x=cos^3t y=sin^3t.....find the cartesian equation with the parametric equations.... pls help with this too..

 

[whitecorp]

thanks
x=cos^3t y=sin^3t.....find the cartesian equation with the parametric equations.... pls help with this too..

x= cos^3 t ======> cos^2 t = x^ (2/3) ---------------(1)

y= sin^3 t ======> sin^2 t = y^ (2/3) --------------(2)

(1)+(2): x^(2/3) +y ^(2/3) =1 (shown)

Hope this helps. Peace.

 

[kronix6]

Hi can anyone solve this plz.

use the substitution x=tanQ to show that

( f(1-x^2)/(1+x^2)^2 dx )is equal to (fcos2Q dQ)

i dont know how to make integration sign so i used f instead. so f= integration sign

 

[iKhaled]

Hi can anyone solve this plz.

use the substitution x=tanQ to show that

( f(1-x^2)/(1+x^2)^2 dx )is equal to (fcos2Q dQ)

i dont know how to make integration sign so i used f instead. so f= integration sign

ok bro for this question u need to know trig identities and integration by substitution so here is the solution for it..

x = tan Q

∫ (1-x^2)/ (1+x^2)^2 dx = ∫cos 2Q dQ

dx/dQ = sec^2 Q

dx = sec^2 Q dQ

∫(1-tan^2 Q)/ (1+tan^2 Q)^2 sec^2 dQ

ok i wont add the Q now cuz its a bit annoying will leave that to the end..

we know that 1+tan^2 = sec^2 so
∫ [(1-tan^2)/(sec^2)^2] x sec^2 dQ ( here we substituted the dx with sec^2 Q dQ )
∫(1-tan^2)/sec^2 dQ
∫(1- sin^2/cos^2)/ (1/cos^2) dQ
∫ cos^2 Q - sin^2 Q dQ

we have a rule that says cos^2 Q - sin^2 Q = cos 2Q

therefore.. ∫ cos 2Q dQ

i hope this is clear to u and u understood it now!

 

[ayeshaK]

f(x)=1/6x+5 and g(x)=1/2x+1 show that the equation fg(x) = 1/5 has no roots.

 

[iKhaled]

f(x)=1/6x+5 and g(x)=1/2x+1 show that the equation fg(x) = 1/5 has no roots.

what paper is this or is it 1/6x + 5 and 1/2x + 1 or 1/2x+1 as in 2x+1 the denominator?

 

[Rutzaba]

f(x)=1/6x+5 and g(x)=1/2x+1 show that the equation fg(x) = 1/5 has no roots.

x se x cancel

 

[Rutzaba]

u wud get after simplifying 10x +5 = 11+ 10x

 

[Soldier313]

a2 math



please can someone help

 

[Rutzaba]

a2 math

View attachment 18957


please can someone help

* break e^( 2x+y) into e^2x multiplied by e^y or (e^2x)(e^y)

* take y onto the side of dy and place x onto other side of equation where dx is

Dy (1 / e^y) = dx (e^2x)

* integrate both sides placing a c on the x hand side.
it will give you
Dy (e^-y) -------> -e^ -y
Dx (e^ 2x) ---------> 2e^2x

Thus -e^ -y = 2e^2x +c

* now place values of x n y to take out value of c wch wud b -1.5

* place the value of c into formula = last line of 3rd * above

 

[Rutzaba]

Rearrange and you make y the subject of formula you will get y = ln (-2/ ((e^2x) -3)

 

[Rutzaba]

a2 math

View attachment 18957

please can someone help

Did u understand? Can u chk n tel me if my answers ryt?

 

[Soldier313]

Rearrange and you make y the subject of formula you will get y = ln (-2/ ((e^2x) -3)

Thanx a lot sis!

Did u understand? Can u chk n tel me if my answers ryt?

yup twas right just that ms states the answer as : y = ln(2 /(3 − e 2 x ))
thanx

 

[sma786]

Find the value of the term which is independent of x in the expansion of (x+3/x)^4

And people, does anyone have good notes on math ? Radians, binomial theorm and all, my teacher is the worst teacher and i need some help ! Thankyou <3

 

[Rutzaba]

Find the value of the term which is independent of x in the expansion of (x+3/x)^4

And people, does anyone have good notes on math ? Radians, binomial theorm and all, my teacher is the worst teacher and i need some help ! Thankyou <3

4c2( x^2) ((3/x) ^2)
54

 

[sma786]

4c2( x^2) ((3/x) ^2)
54

but the questions asks the inpendent of x, ur taking x2

 

[Soldier313]

please solve this and explain find the equation of two lines through the origin which are inclined at 45 degree to the line 2y+x=0 please tell me frome where this question is?

http://themediaeducation.com/educat...uestion-cWEtMjAxMjA1MTAwMjE5NThBQUY0ajVu.html

found the qn and soln here.

 

[Rutzaba]

but the questions asks the inpendent of x, ur taking x2

but the questions asks the inpendent of x, ur taking x2

Hunz x^2 in numerator would be cancelled by x^2 in denominator

 

[Rutzaba]

please solve this and explain find the equation of two lines through the origin which are inclined at 45 degree to the line 2y+x=0 please tell me frome where this question is?

did u get the solution? or do wan me to do it for u?