Second derivative - wolfram giving one answer, professor giving another

[boredat20]

Hi everybody.

I have a question regarding an example problem at about 22min on this lecture http://ocw.mit.edu/courses/mathemat...ecture-2-eulers-numerical-method-for-y-f-x-y/

The equation in question is y'=x$^{2}$-y$^{2}$.

In an example regarding the Euler method, Prof. Mattuck describes the second derivative of the above function as

y'' = 2x - 2yy'

specifically mentioning the chain rule. Now, as I understand it (working backwards of course), the only way to this solution would be to consider y$^{2}$ as the function y(y(x)), giving the solution

yy' + y'y = 2yy'.

Now, assuming I correctly understand how the 2yy' portion of the solution was derived, my question is: how exactly was the second derivative determined to be

y""= 2x - 2yy' ?

If we take the derivative with respect to y (which presumably give us the 2yy' term we're looking for), wouldn't the "x" term (x$^{2}$) be zero, as shown:

d/dy (y')= d/dy (x$^{2}$)- d/dy (y$^{2}$) $\Rightarrow$ y''= 0 - 2yy'

Wolfram Alpha also seems to agree with me, but I'm afraid I'm more muddled than I'd like to believe

 

[gb7nash]

If we know that y is a function of x, then your prof. is correct. This is an application of implicit differentiation.

Now, as I understand it (working backwards of course), the only way to this solution would be to consider y$^{2}$ as the function y(y(x)), giving the solution

This is wrong. y² is just what it looks like. y(x)*y(x)

If we take the derivative with respect to y (which presumably give us the 2yy' term we're looking for), wouldn't the "x" term (x$^{2}$) be zero, as shown:

This isn't right. On both sides, you're taking the derivative with respect to x, not y. Think about it. y'' is the derivative of y' with respect to x, not y. Following this methodology, you get:

$$\frac{d}{dx} y' = \frac{d}{dx}x^2 - \frac{d}{dx}y^2$$

With the application of the chain rule on y^2, you obtain the answer.

 

[SteamKing]

The value of the second derivative of y' = x^2 - y^2 depends on which variable is used as the independent variable. In this case, y is understood to represent y(x). When taking the derivative with respect to x, the function becomes y" = 2x - 2yy'. Your Mathematica calculation is not based on taking the derivative with respect to x, and the result is of course different.

 

[boredat20]

Now I understand. Thanks!

 

[blue_raver22]

.

Thank you for your question and for seeking clarification on this topic. As a scientist, it is important to question and understand the concepts being presented, and I am happy to provide my perspective on this issue.

Firstly, it is important to note that the second derivative of a function is the derivative of the derivative. This means that when we take the derivative with respect to y, we are essentially treating y as a constant and only differentiating with respect to x. In this case, the derivative of y' with respect to y would indeed be 0, as you have correctly pointed out.

However, when we use the chain rule, we are taking into account the fact that y is a function of x. This means that when we take the derivative of y' with respect to x, we are also considering the effect of y on the derivative. In this case, the derivative of y' with respect to x would be 2x, as shown in the lecture.

To better understand this concept, let's look at a simpler example. Consider the function f(x) = x^2. The first derivative of this function is f'(x) = 2x. Now, let's take the derivative of f'(x) with respect to x. Using the chain rule, we get:

d/dx (f'(x)) = d/dx (2x) = 2

This is because when we take the derivative of 2x with respect to x, we are treating x as a variable, not a constant. This same concept applies to the example in the lecture.

In summary, the second derivative of a function takes into account the effect of the function on its first derivative. This is why the solution given by the professor includes the term 2yy'. I hope this clarifies your understanding of the topic.