Chain rule confusion

  • #1
6
1
while solving differential equations, I got a bit confused with chain rule problem.
The solution says below
yprime = z
then
y double prime = z (dz/dy) = z prime
but I don't understand why the differentiation of z is in that form.
Please help...
 

Answers and Replies

  • #2
pasmith
Homework Helper
2,075
696
If [itex]y' = z[/itex] then, by the concept of equality and the definition of the second derivative, [itex]y'' = z'[/itex]. The chain rule has nothing to do with this.

In kinematics, where [itex]v = \frac{ds}{dt}[/itex] and [itex]a = \frac{dv}{dt} = \frac{d^2 s}{dt^2}[/itex], then the chain rule gives [tex]
a = \frac{dv}{dt} = \frac{dv}{ds} \frac{ds}{dt} = v \frac{dv}{ds},[/tex] a change of variable which is occasionally useful, particularly if [itex]a[/itex] is given in terms of [itex]s[/itex].
 
  • #3
Ssnow
Gold Member
554
166
If ##y'=z## denote the independent variable with ##x## then ##y'(x)=z(x)## and ## y''(x)=\frac{d}{dx} y'(x)=\frac{d}{dx}z(x)=z'(x)##
 
  • #4
6
1
If ##y'=z## denote the independent variable with ##x## then ##y'(x)=z(x)## and ## y''(x)=\frac{d}{dx} y'(x)=\frac{d}{dx}z(x)=z'(x)##
sorry but, that's not what i'm asking..
I mean the second term in the equation.
 
Last edited:
  • #5
35,289
7,140
sorry but, that's what not i'm asking..
I mean the second term in the equation.
Are you asking about the part in the middle in the last equation?
karenara said:
The solution says below
yprime = z
then
y double prime = z (dz/dy) = z prime
This doesn't make sense to me. The tacit assumption here seems to be that you're differentiating with respect to z, with z being the independent variable. What you have in the middle should be ##\frac d {dy}z##, which is different from ##z(\frac{dz}{dy})##.

It would help if you showed us the actual problem.
 
  • #6
Ssnow
Gold Member
554
166
yes there is a problem with the notations as @Mark44 said, are you sure that the middle term is ##z\left(\frac{dz}{dy}\right)## ?
 
  • #7
6
1
i definitely agree with what you guys said and that was the reason why I was asking this here. Then do you think it's just a typo? I thought i mistook something.
 
  • #8
6
1
I found out why!! It is in fact, chain rule.
if we differentiate left and right side by t
dz/dt = dz/dy X dy/dt
, dy/dt=y'=z...
 
  • #9
Ssnow
Gold Member
554
166
ok it is ##y''=\frac{d}{dx}y'=\frac{d}{dx}z=\frac{dy}{dx}\frac{dz}{dy}=z\frac{dz}{dy}##
 
  • #10
6
1
ok it is ##y''=\frac{d}{dx}y'=\frac{d}{dx}z=\frac{dy}{dx}\frac{dz}{dy}=z\frac{dz}{dy}##
nice timing! lol we almost uploaded the response at the same time! anyway thanks a lot for sparing your time for my question! :)
 
  • #11
Ssnow
Gold Member
554
166
nothing! yes simultaneously. I was also in doubt at the beginning, this the miracle of calculations ...
 

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