If `e^y(x+1)=1` . Show that `(d^2y)/(dx^2)=((dy)/(dx))^2`

Explanation

`e^y(x+1)=1`

Differentiating w.r.t. x

`(d(e^y(x+1)))/dx=(d(1))/dx`

`(d(e^y(x+1)))/dx=0`

Using Product Rule

`(d(e^y))/dx.(x+1)+(d(x+1))/dx.e^y=0`

`(d(e^y))/dxxxdy/dx(x+1)+((d(x))/dx+(d(1))/dx).e^y=0`

`(d(e^y))/dyxxdy/dx(x+1)+(1+0).e^y=0`

`e^yxxdy/dx(x+1)+e^y=0`

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

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

`dy/dx=(-1)/((x+1))`

Again Differentiating w.r.t. x

`d/dx(dy/dx)=d/dx((-1)/((x+1)))`

Using Quotient Rule

`(d^2y)/(dx^2)=-[((d(1))/dx.(x+1)-(d(x+1))/dx.1)/(x+1)^2]`

`(d^2y)/(dx^2)=-[(0.(x+1)-(d(x+1))/dx.1)/(x+1)^2]`

`(d^2y)/(dx^2)=-[(0-(1+0).1)/(x+1)^2]`

`(d^2y)/(dx^2)=-[(-1)/(x+1)^2]`

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

Hence

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

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

`(d^2y)/(dx^2)=((dy)/(dx))^2`

Hence Proved