Undergrad How to Differentiate Using the Chain Rule?

[tomwilliam]

I'm coming back to maths (calculus of variations) after a long hiatus, and am a little rusty. I can't remember how to do the following derivative:

$\frac{d}{d\epsilon}\left(\sqrt{1 + (y' + \epsilon g')^2}\right)$
where $y, g$ are functions of $x$

I know I should substitute say $u = 1 + (y' + \epsilon g')^2$
then use the chain rule, $\frac{\partial\sqrt{u}}{\partial x} \frac{\partial x}{\partial \epsilon}$

But now I'm a little stuck. Can anyone help with a pointer?
I know what the final answer is, but can't get there.
Thanks

 

[PeroK]

I'm coming back to maths (calculus of variations) after a long hiatus, and am a little rusty. I can't remember how to do the following derivative:

$\frac{d}{d\epsilon}\left(\sqrt{1 + (y' + \epsilon g')^2}\right)$
where $y, g$ are functions of $x$

I know I should substitute say $u = 1 + (y' + \epsilon g')^2$
then use the chain rule, $\frac{\partial\sqrt{u}}{\partial x} \frac{\partial x}{\partial \epsilon}$

But now I'm a little stuck. Can anyone help with a pointer?
I know what the final answer is, but can't get there.
Thanks

You have a function $f(u(\epsilon))$ and the chain rule is:
$$\frac{df}{d\epsilon} = \frac{df}{du}\frac{du}{d\epsilon}$$ Note that $x$ and$y$ don't come into this.

 

[tomwilliam]

Ok, thanks, I see where my mistake came in.
So now I have
$\frac{d\sqrt{u}}{du}=(1/2)u^{-1/2}$
and
$\frac{du}{d\epsilon}= 2(y' + \epsilon g')$

so $\frac {df}{d\epsilon} = (1/2)u^{-1/2}\times 2(y' + \epsilon g')$

but I seem to be a factor of $g'$ out.

 

[PeroK]

$\frac{du}{d\epsilon}= 2(y' + \epsilon g')$

You need to take more care over that step.

 

[Mark44]

I can't remember how to do the following derivative:

$\frac{d}{d\epsilon}\left(\sqrt{1 + (y' + \epsilon g')^2}\right)$
where $y, g$ are functions of $x$

I know I should substitute say $u = 1 + (y' + \epsilon g')^2$
then use the chain rule, $\frac{\partial\sqrt{u}}{\partial x} \frac{\partial x}{\partial \epsilon}$

Since it's not given that $\epsilon$ is a function of x, there is no need for partial derivatives here. As @PeroK already noted, y and g (and therefore y' and g') don't enter into the calculation at all. They can all be considered to be constants, as far as the differentiation goes.
With your substitution, $u = 1 + (y' + \epsilon g')^2 = h(u)$,
the chain rule would look like this: $\frac d {du}(u^{1/2}) \cdot \frac {du}{d\epsilon}$.

 

[tomwilliam]

Thanks to both of you...I understand it now.