Solve Chain Rule Confusion with Diff. Eq. | Help

[karenara]

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...

 

[pasmith]

If $y' = z$ then, by the concept of equality and the definition of the second derivative, $y'' = z'$. The chain rule has nothing to do with this.

In kinematics, where $v = \frac{ds}{dt}$ and $a = \frac{dv}{dt} = \frac{d^2 s}{dt^2}$, then the chain rule gives $$a = \frac{dv}{dt} = \frac{dv}{ds} \frac{ds}{dt} = v \frac{dv}{ds},$$ a change of variable which is occasionally useful, particularly if $a$ is given in terms of $s$.

 

[Ssnow]

If $y'=z$ denote the independent variable with $x$ then $y'(x)=z(x)$ and $y''(x)=\frac{d}{dx} y'(x)=\frac{d}{dx}z(x)=z'(x)$

 

[karenara]

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.

 

[Mark44]

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?

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.

 

[Ssnow]

yes there is a problem with the notations as @Mark44 said, are you sure that the middle term is $z\left(\frac{dz}{dy}\right)$ ?

 

[karenara]

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.

 

[karenara]

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...

 

[Ssnow]

ok it is $y''=\frac{d}{dx}y'=\frac{d}{dx}z=\frac{dy}{dx}\frac{dz}{dy}=z\frac{dz}{dy}$

 

[karenara]

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! :)

 

[Ssnow]

nothing! yes simultaneously. I was also in doubt at the beginning, this the miracle of calculations ...