What is the Second Derivative of 2xy - y^3 = 5?

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In summary, the conversation is about a homework problem where the student is attempting to find the second derivative of a function. They provide their work and ask for help in identifying their mistake. The expert summarizes the conversation by pointing out where the mistake occurred and providing a hint. The student then asks for clarification on the function and its variables. The expert clarifies and provides guidance on how to properly approach the problem. In the end, the conversation is about implicit differentiation and finding the derivatives of functions of two variables.
  • #1
LearninDaMath
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Homework Statement



Here is my work/attempt. In blue is my answer. In green is the supposed correct answer.

The first derivative is surely correct, as it matches the answer I was supposed to get. However, if my second derivative answer is incorrect, can you determine where I I am making a mistake?


doubleprimeimplicit-1-1.jpg
 
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  • #2
Here's a hint: your mistake comes on the third to last step (the line beginning f''(x) = -2(-2y)
 
  • #3
tjackson3 said:
Here's a hint: your mistake comes on the third to last step (the line beginning f''(x) = -2(-2y)

Would it be correct to approach this step like this:

doubleimplicit2-1.jpg
EDIT: On the first line, please regard the 12y^2 as just 12y. The second line is written as though the 12y is what I started off with.
 
Last edited:
  • #4
Your mistake is in the 3rd to last step, where you have one term with a 2x-3y^2 denominator. In the following step you simplified the equation by multiplying (2x-3y^2) with the (2x-3y^2)^2 term. However the two 4y terms on the numerator were not being divided by 2x-3y^2.
 
  • #5
Would this be the correct way to do that step?


doubleimplicit3-1.jpg
 
  • #6
Okay, I was able to get the answer in the right form as the provided answer. However, it is off numerically by 2. The provided answer has 18xy and my answer has 16xy. I have always known 8 + 8 to equal 16...what gives?


doubleimplicit4.jpg
 
  • #7
Some comments.
1) You don't take "f' " of an expression. f'(x) is the derivative of f. The operator is d/dx. f' and f'' should not appear in your work at all.
2) You wrote f(x) = 2xy - y3 = 0. This is incorrect because a) the expression 2xy - y3 involves both x and y, not just x alone, and b) the equation 2xy - y3 = 0 defines an implicit relationship between x and y.
 
  • #8
Ah, that's right. It's not a function is it. It's an equation, but not a function. Appreciate the input.
 
  • #9
As Mark44 said, it's an implicit relationship. Correct me if I'm wrong, but doesn't the question seem a little strange? How can

[itex]f(x) = 2xy - y^{3} = 5[/itex]?

That being the case, wouldn't we need to write

[itex]f(x,y) = 2xy - y^{3} = 5[/itex], or [itex]f(x,y) = 2xy - y^{3} - 5[/itex]

since it is a function of two variables? Then we would just need to find [itex]\frac{\partial f}{\partial x}[/itex] and [itex]\frac{\partial f}{\partial y}[/itex]. If this isn't the case, we would just have

[itex]2xy - y^{3} = 5[/itex],

and we would need to find [itex]\frac{dy}{dx}[/itex] implicitly, not f'(x).

So guys, which is it? Or am I completely off on this one?
 
  • #10
stripes said:
As Mark44 said, it's an implicit relationship. Correct me if I'm wrong, but doesn't the question seem a little strange? How can

[itex]f(x) = 2xy - y^{3} = 5[/itex]?

That being the case, wouldn't we need to write

[itex]f(x,y) = 2xy - y^{3} = 5[/itex], or [itex]f(x,y) = 2xy - y^{3} - 5[/itex]

since it is a function of two variables? Then we would just need to find [itex]\frac{\partial f}{\partial x}[/itex] and [itex]\frac{\partial f}{\partial y}[/itex]. If this isn't the case, we would just have

[itex]2xy - y^{3} = 5[/itex],

and we would need to find [itex]\frac{dy}{dx}[/itex] implicitly, not f'(x).

So guys, which is it? Or am I completely off on this one?

No, I think you are on track here. f'(x) is meaningless here, since the left side of the equation 2xy - y3 - 5 = 0 represents a function of two variables.

The tacit assumption here, I believe, is that y is an implicit function of x, and goal is to find dy/dx using implicit differentiation. Note that this should be written as dy/dx = ..., not f'(x) = ..., so as to not cause confusion between the function (of two variables) in the first equation, and the function (of one variable) that relates x and y.

After dy/dx has been found, differentiate again to find d2y/dx2.
 

1. What is the derivative of 2xy - y^3?

The derivative of 2xy - y^3 is 2y - 3y^2.

2. How do you find the value of "F''(x) of 2xy - y^3 = 5" at a specific point?

To find the value of "F''(x) of 2xy - y^3 = 5" at a specific point, you can plug in the x and y values into the derivative equation 2y - 3y^2. The resulting number will be the value of the second derivative at that point.

3. What does the second derivative of 2xy - y^3 represent?

The second derivative of 2xy - y^3 represents the rate of change of the slope of the original function. In other words, it shows how the slope of the original function is changing at a specific point.

4. Can the second derivative of 2xy - y^3 be negative?

Yes, the second derivative of 2xy - y^3 can be negative. This indicates that the slope of the original function is decreasing at that particular point.

5. How does the second derivative of 2xy - y^3 relate to the concavity of the function?

The second derivative of 2xy - y^3 can determine the concavity of the function. If the second derivative is positive, the function is concave up, and if it is negative, the function is concave down. A second derivative of zero indicates a point of inflection.

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