Question

How do you find the slope of a tangent line to the graph of the function `3xy-2x+3y^2=5` at (2,1)?

Answer 1

Please the explanation for steps leading to the slope, `m = -1/12`

Explanation:

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

The steps are:
1. Implicitly differentiate the equation.
2. Obtain the slope of the tangent line by evaluating the derivative at the point `(2, 1)`

1. Implicitly differentiate the equation.

I will differentiate each term, separately.

1.1 Use the product rule for term 1:

`(uv)' = u'v + uv'`

let `u = 3x`, then `u' = 3, v = y and v' = dy/dx`

`(uv)' = 3y + 3xdy/dx`

1.2 Use the power rule on term 2:

`(d(-2x))/dx = -2`

1.3 Use the chain rule on term 3:

`(d(3y^2))/dx = 6ydy/dx`

1.4 Term 4 is a constant, 5, the derivative is zero.

Put these back into the equation:

`3y + 3xdy/dx - 2 + 6ydy/dx = 0`

Move the terms without `dy/dx` to the right:

`3xdy/dx + 6ydy/dx = 2 - 3y`

Factor out `dy/dx` on the left:

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

Divide both sides by `(3x + 6y)`:

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

The slope, m, of the tangent line is the above evaluated at the point `(2,1)`:

`m = (2 - 3(1))/(3(2) + 6(1))`

`m = -1/12`