determine the first derivatives for the function?

Question

f(x)=(1+x)^2 / 1 - x

Answer 1

The function is of the form d(u/v) which is given by
(v du - u dv) / v^2.
here u=(1+x)^2, hence du=2(1+x)
here v=1-x, hence dv=-1.
substitute these values in the formula and simplify to get final result.

KSR

Answer 2

do u want it be solved by first principle?
let dx = delta x, small change in x

then first find f(x+dx)
f(x +dx) = (1 + x + dx)^2/(1 - x -dx)

Now (1+x + dx)^2 = (1+x)^2 + 2(1+x)dx +(dx)^2

and therefore
f(x+dx) = [(1+x)^2 +2(1+x)dx +(dx)^2]/(1-x-dx)
first derivative is given by Limit (dx-->0)[ f(x+dx)-f(x)]/dx

calculate f(x+dx)-f(x)
= {(1+x)^2 +2(1+x).dx +(dx)^2]/(1-x-dx)} - {(1+x)^2 / (1 - x)}

= [1/(1-x)].{ [ {(1+x)^2 +2(1+x)dx +(dx)^2} / {1-(dx/(1-x)} ] - [(1+x)^2] }
=[1/(1-x)].{ [ (1+x)^2 +2(1+x)dx +(dx)^2 ] - [(1+x)^2].[1-(dx/((1-x))]}


=[1/(1-x)]. { (1+x)^2 +2(1+x)dx +(dx)^2 - (1+x)^2+ (dx(1+x)^2/((1-x)) }
==[1/(1-x)]. { +2(1+x)dx +(dx)^2 + (dx(1+x)^2/((1-x)) }
now divide by dx
so (f(x+dx)-f(x))/dx=[1/(1-x)]. { +2(1+x) +(dx) + ((1+x)^2/((1-x)) }
in the limit dx-->0
this becomes df(x)/dx or f'(x) =[1/(1-x)]. { +2(1+x) + ((1+x)^2/((1-x)) }


hence the first derivTIVE OBTAINED

Answer 3

i'm assuming: (1+x)^2/(1-x)
then:
( 2(x+1)(1-x)-(-1)(1+x)^2)/(1-x)^2=
(x+1)(3-x)/(1-x)^2
easy and simple as it can get

Answer 4

f(x) = g(x)/h(x) so
f'(x) = (g'(x)h(x) - h'(x)g(x))/(h(x))²

f(x) = (g(x))^n so
f'(x) = (g(x))^(n-1) * g'(x)

f(x) = g(x) + h(x) so
f'(x) = g'(x) + h'(x)

Now get off yer dead αss and go do your homework ☺


Doug

Answer 5

f'(x)=(1-x)*2(1+x)-(1+x)^2(-1)/(1-x)^2
2(1-x^2)+(1+x)^2/(1-x)^2
=2-2x^2+1+x^2+2x/(1-x)^2
=-(x^2-2x-3)/(1-x)^2
=-(x-3)(x+1)/(1-x)^2