Question

Find y' ?? please help

Answer 1

y'=`[24x^5 e^[2y]-8x]/[1-8x^6e^[2y]`]

Explanation:

`sqrt[4x^2+y]`.... =`4x^6e^[2y]` squaring both sides will give
`4x^2+y`=`4x^6e^[2y]`........Differentiating both sides
implicitly with respect `x` and using the product rule on the right hand side, i.e, `d[uv]=vdu+udv`

Let `u =4x^6 and` `v=e^[2y]`

`8x+dy/dx`=`e^[2y]d/dx4x^6`+`4x^6d/dxe^[2y]`

So , `8x+dy/dx`=`e^[2y]` `24x^5`+`4x^6` `e^[2y]` `.2dy/dx` ,rearranging and factoring........

`dy/dx-8x^6e^[2y]dy/dx`=`24x^5e^[2y]-8x`

`dy/dx[1-8x^6e^[2y]]`=`24x^5e^[2y]-8x` and so,

`y'`=`[24x^5e^[2y]-8x]/[1-8x^6e^[2y]`

Answer 2

`dy/dx = ( 24x^5e^2y -8x) / ( 1- 8x^6e^(2y) )`

Explanation:

We have:

`sqrt(4x^2+y) = 2x^3e^y`

If we have a function `F(x, y) = 0` which defines `y` as a function of `x, y = y(x)`, then:

`dy/dx = − F_x / F_y`

So Let `F(x,y) = sqrt(4x^2+y)-2x^3e^y`; Then;

Then:

`F_x = (partial F)/(partial x)`

`\ \ \ \ = 1/2(4x^2+y)^(-1/2)(8x)-6x^2e^y`

`\ \ \ \ = (4x)/sqrt(4x^2+y)-6x^2e^y`

And:

`F_y = (partial F)/(partial y)`

`\ \ \ \ = 1/2(4x^2+y)^(-1/2)(1)-2x^3e^y`

`\ \ \ \ = 1/(2sqrt(4x^2+y))-2x^3e^y`

And so we have:

`dy/dx = − F_x / F_y`

`\ \ \ \ \ = − ((4x)/sqrt(4x^2+y)-6x^2e^y) / (1/(2sqrt(4x^2+y))-2x^3e^y)`

Using the original definition:

`sqrt(4x^2+y) = 2x^3e^y`

We can simplify our expression:

`dy/dx = − ((4x)/(2x^3e^y)-6x^2e^y) / (1/(2(2x^3e^y))-2x^3e^y)`

`\ \ \ \ \ = − ( ( 4x -6x^2e^y(2x^3e^y) ) / (2x^3e^y) ) / ( ( 1-2x^3e^y(2)(2x^3e^y) ) / ((2)(2x^3e^y) )`

`\ \ \ \ \ = −2 ( 4x -6x^2e^y(2x^3e^y) ) / ( 1-2x^3e^y(4x^3e^y) )`

`\ \ \ \ \ = ( 24x^5e^2y -8x) / ( 1- 8x^6e^(2y) )`