Question

How do you find `(dy)/(dx)` given `(2y-x)^2+3x=0`?

Answer 1

Differentiate each term:

`(d((2y-x)^2))/dx + (d(3x))/dx = (d(0))/dx" [1]"`

The first term requires the use of the chain rule:

`(d(f(g(x))))/dx = (df)/(dg)(dg)/dx" [2]"`

Let `g = 2y - x`, then, `f(g) = g^2`, `(df)/(dg) = 2g`, and `(dg)/dx =2dy/dx-1`

Substituting into equation [2]:

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

Reverse the substitution for g:

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

Substitute back into equation [1]:

`2(2y - x)(2dy/dx-1) + (d(3x))/dx = (d(0))/dx`

The derivative of the second term is trivial and the derivative of 0 is 0:

`2(2y - x)(2dy/dx-1) + 3 = 0`

Solve for `dy/dx`:

`2(2y - x)(2dy/dx-1) = -3`

`2dy/dx-1 = -3/(2(2y - x))`

`2dy/dx=1 -3/(2(2y - x))`

`dy/dx=1/2 -3/(4(2y - x))`