Taylor/Maclaurin Differentiation explanation needed/Differentiation from y' to y''

[FranzGans]

I need to do some differentiation for a Taylor Maclaurin series.
I have two questions in specific.

I have got y''=y'ln(y)+x given
then I have to differentiate again and it says:
y'''= y''ln(y)+(y')²(y^-1)+1

Why is this not

y'''=y''ln(y)+y'y^-1+1

Where and why did the extra y' in the middle bit come from?

the next bit he then gives as:

y^4= y'''ln(y)+y''y'(y^-1)+2y'y''(y^-1)+(y')²(-y^-2)y'

which then confused me, for if i would have gotten the correct result for y''' my answer would have been ( and I have no idea where the extra y' and y'' came from)

y^4=y'''ln(y)+y''(y^-1)+y''(y^-1)-y'(y^-2)

also there is an example where it says:

y''=-sin(y)

which gets differentiated to be

y'''=-cos(y)y'

Again there is a y' appearing and I have no idea where from.

then again differentiating this again it gives

y^4=sin(y)(y')²-cos(y)y''

again, if I had the correct guess on y''' my answer for this would have been

y^4=sin(y)y'-cos(y)y''

there are a lot of y's appearing and I have no clue why that is. I must be missing a certain rule but I could not find it after 2 hours, so I am starting to get frustrated.

Thank you guys for your time and help.

 

[Deleted member 4993]

I need to do some differentiation for a Taylor Maclaurin series.
I have two questions in specific.

I have got y''=y'ln(y)+x given
then I have to differentiate again and it says:
y'''= y''ln(y)+(y')²(y^-1)+1

Why is this not

y'''=y''ln(y)+y'y^-1+1

Where and why did the extra y' in the middle bit come from?

the next bit he then gives as:

y^4= y'''ln(y)+y''y'(y^-1)+2y'y''(y^-1)+(y')²(-y^-2)y'

which then confused me, for if i would have gotten the correct result for y''' my answer would have been ( and I have no idea where the extra y' and y'' came from)

y^4=y'''ln(y)+y''(y^-1)+y''(y^-1)-y'(y^-2)

also there is an example where it says:

y''=-sin(y)

which gets differentiated to be

y'''=-cos(y)y'

Again there is a y' appearing and I have no idea where from.

then again differentiating this again it gives

y^4=sin(y)(y')²-cos(y)y''

again, if I had the correct guess on y''' my answer for this would have been

y^4=sin(y)y'-cos(y)y''

there are a lot of y's appearing and I have no clue why that is. I must be missing a certain rule but I could not find it after 2 hours, so I am starting to get frustrated.

Thank you guys for your time and help.

All these have to do with chain rule:

u = f[y(x)]

u' = df/dy * dy/dx

If you keep this in mind - all your questions will be answered.

Let's start with:

u = ln[y(x)]

then

du/dx = du/dy * dy/dx = 1/y * y'

Another problem:

u = cos[y(x)]

u' = -sin[y(x)] * y'

That's it.....

 

[FranzGans]

Sorry, but it's been a long day for me and I do not quite follow, and the things I just read on Wikipedia do not make too much sense anymore.

Could you rewrite this in the same form as I used?

e.g. y'''= because...

Thanks a lot though, its just not my time anymore...

 

[Deleted member 4993]

Sorry, but it's been a long day for me and I do not quite follow, and the things I just read on Wikipedia do not make too much sense anymore.

Could you rewrite this in the same form as I used?

e.g. y'''= because...

Thanks a lot though, its just not my time anymore...

Do you understand how chain rule is here?

Did you understand the examples I had presented?

 

[FranzGans]

Unfortunately not. I do think I understand the chain rule but I fail to apply it, especially for
y=-sin(y)
y''=-cos(y)y'

 

[Deleted member 4993]

Unfortunately not. I do think I understand the chain rule but I fail to apply it, especially for
y=-sin(y)
y''=-cos(y)y'

.
Is that y=-sin(y)


That is:

Do you have y on both sides of the equality sign?

Your original problem was:

y''=-sin(y)

which gets differentiated to be

y'''=-cos(y)y'

which is quite different.

 

[FranzGans]

My bad I wanted to write it as
y'=-sin(y)
y''=-cos(y)y'

or would that be different if it was y'' and y'''?

I think I understand the basic example of the chain rule from that link, but I just cannot apply it to my problems.