I need to find the derivitive of y=2x/(1-3x^2)?

Answer 1

Quotient rule:

dy/dx = (vu' - uv')/v^2 where u = 2x, v= (1-3x^2)

therefore
dy/dx = [(1 - 3x^2)(2) -(2x)(-6x)]/(1 - 3x^2)^2
= [2-6x^2 +12x^2]/(1-3x^2)^2
=(6x^2 +2)/(1 - 3x^2)^2

Answer 2

this is the quotient rule, which is the denomenator times the derivative of the numerator, minus the numerator times the derivative of the denomenator, all over the denomenator squared.

so you have:

y prime = (1-3x^2) * 2 - 2x * (-6x)
all over (1-3x^2)^2

you can simplify this as:

2 [ (1-3x^2) + 6x] over (1-3x^2)^2

i'm not sure it can be simplified any more though...

if you have any more questions email me !

Answer 3

If you like the SIMPLE answer, use this link:
http://wims.unice.fr/wims/wims.cgi

Here is the answer:
f(x) = 2x/(1-3x^2)
f'(x) = ( 2 / (1-3x^2) ) + ( 12x / (1-3x^2)^2 )

Answer 4

Use the quotient rule:

(f' * g - f * g')/g^2

Let the numerator 2x be f.
Let the denominator (1-3x^2) be g

y' = ((2 * (1-3x^2)) - (2x * (-6x))/(1-3x^2)^2

And of course, simplify.

Answer 5

y = u / v
u = 2 x
v = 1 - 3 x ²
dy/dx = (v du/dx - u dv/dx) / v ²
du/dx = 2
dv/dx = - 6x
dy/dx = [ (1 - 3 x ²)(2) - (2x)(- 6x) ] / (1 - 3 x ²) ²
dy/dx = [ 2 - 6 x ² + 12 x ² ] / ( 1 - 3 x ² ) ²
dy/dx = [2 - 6 x ²] / (1 - 3 x ²) ²
dy/dx = 2 ( 1 - 3 x ² ) / (1 - 3 x ² ) ²
dy/dx = 2 / ( 1 - 3 x ² )

Answer 6

u shld be able to apply ur QUOtient rule. and pls solve dis urself using the formula. dun ask others for the answers cos dis is math. u nid practice. :)

Answer 7

d (f (x) / g (x)) = ( g df(x) - f dg(x) )/( (g (x))^2 )