differential equation and series

[Sonal7]

I am really not sure how they have worked out the d^3y/dx^3, I think that the last two terms of that line is the differential of the y^2 cos x, but i dont understand how one gets 3 terms for the first term. Is this series using Leibniz's theorem Sorry if this is a rubbish q.

 

[Deleted member 4993]

View attachment 18783

I am really not sure how they have worked out the d^3y/dx^3, I think that the last two terms of that line is the differential of the y^2 cos x, but i dont understand how one gets 3 terms for the first term. Is this series using Leibniz's theorem Sorry if this is a rubbish q.

No Liebniz here - just straight-forward differentiation using chain rule and product rule.

There are three components of the first term [y, dy/dx and sin(x)]. Differentiation, will produce three terms, using product rule. I don't see the confusion here. Are you using pencil/paper while working with this problem or are you just staring at the screen?!

 

[Sonal7]

I am def using pen and paper. I am just not used to doing product rule and you get 3 terms. I have always ended up with 2 terms. Except when you use binomial expansion type of problem. I will think it over.

 

[Sonal7]

I get this now, there are three terms uvw. The product gives 3 terms. thanks, I had never seen this before.
Thanks you !

 

[Steven G]

Consider y = f(x)g(x)h(x)
Now you only know how to compute the derivative for a product of two functions. So think of y as y = (f(x)g(x))h(x). That is, think of f(x)g(x) as just one function (until you want to find the derivative of it!). After all if you actually multiply f(x) by g(x) you do get one function. Now apply the product rule.

So y' = (f(x)g(x))' h(x) + (f(x)g(x)) h'(x) = (f'(x)g(x) + f(x)g'(x)) h(x) + f(x)g(x)h'(x) = f'(x)g(x)h(x) + f(x)g'(x)h(x) + f(x)g(x)h'(x)

You need to make the product rule work for you!

 

[Sonal7]

This is very helpful.