Question

Question #d5dc1

Answer 1

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

Explanation:

This is a textbook case of implicit differentiation. We differentiate both sides of the equation with respect to `x` and treat `y` as a function of `x`. After that all we need to do is solve for the derivative.

We start by taking the derivative of both sides. On the left we'll have to use the product rule:
`d/dx(3xy+y)=d/dx(3x)`

`d/dx(3xy)+dy/dx=3`

`d/dx(3x)*y+3x*dy/dx+dy/dx=3`

`3y+3x*dy/dx+dy/dx=3`

Next we want to solve for `dy/dx`. First we'll bring all of the terms which don't have a `dy/dx` to the other side:
`3x*dy/dx+dy/dx=3-3y`

Next we can factor out a `dy/dx` on the left:
`dy/dx*(3x+1)=3-3y`

And then we can divide both sides by `3x+1`:
`dy/dx*cancel((3x+1))/cancel(3x+1)=(3-3y)/(3x+1)`

`dy/dx=(3-3y)/(3x+1)`
which is our answer.

Answer 2

`dy/dx=3/(3x+1)^2`.

Explanation:

Given that, `3xy+y=3x`.

`:. y(3x+1)=3x`.

`:. y=(3x)/(3x+1)`.

Applying the Quotient Rule, we get,

`dy/dx=3d/dx{x/(3x+1)}`,

`=3*{(3x+1)d/dx(x)-xd/dx(3x+1)}/(3x+1)^2`,

`=3*{(3x+1)*1-x(3)}/(3x+1)^2`.

`rArr dy/dx=3/(3x+1)^2`.