Question

How do you differentiate `4xy-3x-11=0`?

Answer 1

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

Explanation:

Assume the equation `4xy-3x-11=0` implicitly defines `y` as a function of `x` and then differentiate with respect to `x`, using the Product Rule:

`4y+4x dy/dx-3=0`

Now solve for `dy/dx` to get `dy/dx=(3-4y)/(4x)`.

If you happen to know a specific point `(x,y)` on the curve defined by the original equation `4xy-3x-11=0`, you can plug the coordinates of that point into `dy/dx=(3-4y)/(4x)` to find the slope of the curve at that point.

The particular equation `4xy-3x-11=0` is simple enough that we can check our work another way; we can solve for `y` explictly as a function of `x`:

`4xy-3x-11=0\Rightarrow y=(3x+11)/(4x)=3/4+11/4 x^{-1}`

Now we can differentiate this in the ordinary way to get `dy/dx=-11/4 x^{-2}=-11/(4x^2)`. Is this answer the same as the original? It sure looks different. It can be seen to be the same answer by substituting `y=(3x+11)/(4x)` into `dy/dx=(3-4y)/(4x)` in place of `y` and simplifying:

`dy/dx=(3-4((3x+11)/(4x)))/(4x)=(3x-(3x+11))/(4x^2)=-11/(4x^2)`

It is the same!

As an example of a point on this curve, we can use the equation `y=(3x+11)/(4x)` to find `y` when `x=1` to be `y=14/4=7/2`, meaning that the point `(x,y)=(1,7/2)` is on the graph of `4xy-3x-11=0`. The slope of the curve at that point is `dy/dx=(3-14)/4=-11/4=-2.75`.

The equation of the tangent line to the curve at that point is therefore `y=-11/4(x-1)+7/2=-11/4 x+25/4`.

Here's a graph of the situation just described:

Answer 2

Assuming that we want to find `dy/dx`:

Using implicit differentiation, we get: `dy/dx = (3-4y)/(4x)`.

Solving for `y` first, we get: `dy/dx = 11/(4x^2)`.

Explanation:

The question is posted under "Implicit Differentiation", so let's do it that way first:

`4xy-3x-11=0`

Leaving the function Implicit
In order to differentiate `4xy`, we will need the product rule.

Remember that we are assuming that `y` is some function of `x`, so we have `4xy = 4xf(x)` and we use the product rule to get:
the derivative is: `4f(x)+4xf'(x)`

Back to this problem:

`d/dx(4xy)-d/dx(3x)-d/dx(11)=d/dx(0)`

`4y+4xdy/dx-3=0`

`4xdy/dx = 3-4y`

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

Making the function explicit

Solve `4xy-3x-11=0` for `y`

`y = (3x+11)/(4x)`

We could differentiate using the quotient rule, but it is perhaps simpler to rewrite again:

`y = (3x)/(4x)+11/(4x)`

`= 3/4 +11/4x^-1`

So
`dy/dx = -11/4x^-2`

`= -11/(4x^2)`

The answers are equivalent

To see that the answer are equivalent compare:

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

with

`y = (3x+11)/(4x)` and `dy/dx = -11/(4x^2)`

Using Implicit differentiation, there is still a `y` in the derivative. That is the price we pay for not making the function explicit before differentiating. If we substitute the solution for `y`, we get:

`dy/dx = (3-4((3x+11)/(4x)))/(4x)`

`= (3-(3x+11)/x)/(4x)`

`= (3x-3x-11)/(4x^2)`

`= -11/(4x^2)`