How to find slope of tangent line to curve 2xy - y^3 = 4 at (3, 2)?
In this instruction, we will learn techniques for implicit differentiation. The question here is to find the slope of the tangent to 2xy - y^3 - 4 at the point (3, 2). Now, for this function, it is actually difficult to isolate y and explicitly write an equation of y in terms of x. To find the slope, we need to find the derivative, that is dy/dx. We can do implicit differentiation and find dy/dx.
We are given 2xy - y^3 = 4. Let us differentiate both sides with respect to x. That is to say, we'll do d/dx of 2xy - d/dx of y^3 = d/dx of 4. Now, the derivative of 2x times y with respect to x, since there are two variables involved, we'll apply the product rule. Correct. So, we can write this as derivative of 2x times y plus 2x times dy/dx. In the second case, we can apply the chain rule. We can write this as d/dy of y^3 times dy/dx. Great. Here, we know the derivative of a constant is 0. Perfect. Now, we can combine or rather factor dy/dx and then find what dy/dx is.
The derivative of 2x is 2, so we get 2y in the first case, plus 2x dy/dx. Minus the derivative of y^3 with respect to y will be 3y^2, and we have d/dx equal to 0. Now, you can take dy/dx common and take 2y to the other side. Correct. So, we can write this as dy/dx times (2x - 3y^2) = 2y. Dividing by this factor, we get dy/dx is equal to 2y / (2x - 3y^2). Correct. That is the derivative.
Now, we need to find the slope of the tangent line at a given point. Right. So, let's do that part. We already found the derivative. Now, let's find the slope at (3, 2). Right. So, dy/dx is equal to (2y) / (2x - 3y^2). We'll just substitute these values: x = 3, y = 2. So, we have (2 * 2) / (2 * 3 - 3 * 2^2). That gives us 4 / (6 - 12). Correct. Or it is 4 / -6, which simplifies to -2 / 3. So, the slope at the given point is -2 / 3. That's our answer.