B Square root differential problem

[knockout_artist]

Hi,

I working on their text this equation did not make sense to me.

From equation 1 it differentiate second term , I wonder how he got second term of equation 2.

What I think is, what I wrote at the bottom

 

[Charles Link]

They are doing a Taylor series and writing $f(x)=f(a)+f'(a)(x-a) +...$ In this case $x=\epsilon$ and $a=0$. $\\$ I agree with their result.

 

[knockout_artist]

but how did they resolve
f(x)=f(a)+f′(a)(x−a)+..
to get second term in eq 2.

 

[Charles Link]

but how did they resolve
f(x)=f(a)+f′(a)(x−a)+..
to get second term in eq 2.

The chain rule. In numerator of $f'(\epsilon)$ you have $2(y'+\epsilon h) h$. You then compute $f'(0)$.

 

[knockout_artist]

but then why there is no ϵh' term in denominator in second term in equation 2 ?
It should have stayed there.

and if were to consider ϵ = 0 then why there is still e multiplying second term in equation 2 ?

 

[Charles Link]

but then why there is no ϵh' term in denominator in second term in equation 2 ?
It should have stayed there.

and if were to consider ϵ = 0 then why there is still e multiplying second term in equation 2 ?

The $\epsilon$ is from $(x-a)=(\epsilon-0)= \epsilon$. Meanwhile, the $\epsilon h$ in tthe denominator gets put equal to zero as part of taking $f'(0)$. (The entire term is the product of both of these which is $f'(0)(x-a)$.)

 

[knockout_artist]

So like this ?