Question

Find derivative for `x= 1/(1+2r)` , `y=r/(1+r)` ?

Answer 1

`(dy)/(dx)=-1/2((1+2r)/(1+r))^2 or`

`(dy)/(dx)=-1/2(1+y)^2`

Explanation:

We have ,

`x=1/(1+2r) to(1)and y=r/(1+r)to(2)`

Diff. eqn.(1) w.r. to `color(red)(r` we get

#(dx)/(dr)=-1/(1+2r)^2 d/(dr)(1+2r)=color(blue) (-2/(1+2r)^2to(3)#

Again diff. eqn.(2) w.r. to `color(red)( r` we get

`(dy)/(dr)=((1+r)d/(dr)(r)-rd/(dr)(1+r))/(1+r)^2to[because"Quotient Rule"]`

`=>(dy)/(dr)=((1+r)*1-r*1)/(1+r)^2`

`=>(dy)/(dr)=((1+r-r))/(1+r)^2`

`=>(dy)/(dr)=color(blue)(1/(1+r)^2to(4)`

Using `color(blue)((3) and (4)` we get

`(dy)/(dx)=((dy)/(dr))/((dx)/(dr))=[(1/(1+r)^2)/(-2/(1+2r)^2)]`

`=>(dy)/(dx)=-1/2[(1+2r)^2/(1+r)^2]`

`=>color(green)((dy)/(dx)=-1/2((1+2r)/(1+r))^2`

`=>(dy)/(dx)=-1/2(((1+r)+r)/(1+r))^2`

`=>(dy)/(dx)=-1/2(1+r/(1+r))^2`

`to"from (2),we have ": r/(1+r)=y`

`=>color(green)((dy)/(dx)=-1/2(1+y)^2`

Answer 2

`dy/dx=-1/2(1+y)^2`.

Explanation:

Given that, `y=r/(1+r), i.e., y/1=r/(1+r)`.

By componendo-dividendo, `(y+1)/(y-1)={r+(r+1)}/{r-(r+1)}`,

`i.e., (y+1)/(y-1)=(2r+1)/-1`.

`rArr (1-y)/(1+y)=1/(2r+1)`,

`or, x=(1-y)/(1+y)`.

Diff.ing w.r.t. `y,` using the Quotient Rule, we get,

`dx/dy={(1+y)d/dy(1-y)-(1-y)d/dy(1+y)}/(1+y)^2`,

`={(1+y)(-1)-(1-y)(1)}/(1+y)^2`.

`rArr dx/dy=-2/(1+y)^2`.

Consequently, `dy/dx=1/(dx/dy)=-1/2(1+y)^2`, as Respected

Maganbhai P. has readily derived!