Question

Question #bf859

Answer 1

`(d^2y)/dx^2 =(4y^2sec^2(2x)tan(2x)+y^2 -sec^4(2x)-2xsec^2(2x)-x^2) /y^3`

Explanation:

`y^2-x^2=tan2x`

`2y dy/dx-2x=2sec^2(2x)`

`2y dy/dx=2sec^2(2x)+2x`

`dy/dx=(2sec^2(2x)+2x)/(2y)`

`dy/dx=(sec^2(2x)+x)/y`

You can use the quotient rule or just do implicit differentiation again and then substitute in `(sec^2(2x)+x)/y` for `dy/dx`

`(d^2y)/dx^2 =(4sec(2x)sec(2x)tan(2x)+1)/y -(sec^2(2x)+x)/y^2 dy/dx`

`(d^2y)/dx^2 =(4sec(2x)sec(2x)tan(2x)+1)/y -(sec^2(2x)+x)/y^2 *(sec^2(2x)+x)/y`

`(d^2y)/dx^2 =(4sec^2(2x)tan(2x)+1)/y -(sec^4(2x)+2xsec^2(2x)+x^2)/y^3`

`(d^2y)/dx^2 =(4y^2sec^2(2x)tan(2x)+y^2 -sec^4(2x)-2xsec^2(2x)-x^2) /y^3`