Question

How to find `dy/dx` from `x^2y+3xy^2-x=5`?

Answer 1

`dy/dx=(-2xy-3y^2+1)/(x^2+6xy)`

Explanation:

To solve this, we need to use implicit differentiation.

Since we are finding the derivative in respect to `x`, anytime we find the derivative of `y`, we need to tag on a `dy/dx` after.

To start off, we need to use the Product Rule for the first two terms, since there are two variables being multiplied together.

The Product Rule states:

`f'(x)g(x)+f(x)g'(x)`

So we can now set it up like this:

`x^2y+3xy^2-x=5`

`2xy+x^2dy/dx+3y^2+6xydy/dx-1=0`

Now that we have the derivative, we need to clean up the answer a bit. We need to move each term without the `dy/dx` to the other side. So we get:

`x^2dy/dx+6xydy/dx=-2xy-3y^2+1`

Now we factor out a `dy/dx`:

`dy/dx(x^2+6xy)=-2xy-3y^2+1`

Divide:

`dy/dx=(-2xy-3y^2+1)/(x^2+6xy)`

There is your answer

Answer 2

`dy/dx = (1 - 2xy - 3y^2)/(x^2 + 6xy)`

Explanation:

We have:

`x^2y+3xy^2-x=5`

Differentiating wrt `x` by applying the product rule we get:

`x^2(d/dx y) + (d/dx x^2)y + (3x)(d/dx y^2) + (d/dx 3x)y^2 -d/dx x = d/dx 5`

`:. x^2(d/dx y) + (2x)y + (3x)(d/dx y^2) + (3)y^2 -1 = 0`

Then we apply the chain rule to perform the implicit differentiation:

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

Now, we collect terms and solve for `dy/dx`:

`(x^2 + 6xy)(dy/dx) = 1 - 2xy - 3y^2`

`:. dy/dx = (1 - 2xy - 3y^2)/(x^2 + 6xy)`