Question

How do you differentiate this equation?

Question:

If `y=log(sqrtx+1/sqrtx)^2`
then prove that
`x(x+1)^2y_2+(x+1)^2y_1=2`

Answer 1

We will use:
`(1) .ln(M/N)=lnM-lnN`
Quotient rule:
`d/(dx)(u/v)=(v*u^'-u*v^')/(v^2)`

Explanation:

Here,
`y=ln(sqrtx+1/sqrtx)^2=ln((x+1)/sqrtx)^2`
`y=ln((x^2+2x+1)/x)`

`:.y=ln(x^2+2x+1)-lnx`

Diff.w.r.t. `x`,we get

`y_1=1/(x^2+2x+1)(d/(dx)(x^2+2x+1))-1/x`

`y_1=1/(x^2+2x+1)(2x+2)-1/x`

`=>y_1=(2cancel((x+1)))/((x+1)^cancel2)-1/x=2/(x+1)-1/x=(2x-x-1)/(x(x+1))`

`=>y_1=(x-1)/(x(x+1)`

`=>xy_1=(x-1)/(x+1)`

Again diff.w.r.t. `x`, using quotient rule , we get

`xy_2+y_1*1=((x+1)*1-(x-1)*1)/((x+1)^2)=(cancelx+1-cancelx+1)/((x+1)^2`

`=>xy_2+y_1=2/(x+1)^2`

Multiplying each term by `(x+1)^2`

`:.x(x+1)^2y_2+(x+1)^2y_1=2`