# How to Differentiate Using the Chain Rule?

• I
• tomwilliam
In summary, you need to substitute y' and g' into the equation for the derivative, and take more care with the division.
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

tomwilliam said:
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.

Delta2
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.

tomwilliam said:
##
\frac{du}{d\epsilon}= 2(y' + \epsilon g')
##
You need to take more care over that step.

tomwilliam said:
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}##.

Delta2
Thanks to both of you...I understand it now.

## What is the chain rule?

The chain rule is a rule in calculus that allows us to find the derivative of a composite function. It states that the derivative of a composite function is equal to the derivative of the outer function multiplied by the derivative of the inner function.

## When do we use the chain rule?

We use the chain rule when we have a function within another function, also known as a composite function. This commonly occurs when we have nested functions, such as f(g(x)).

## How do we apply the chain rule?

To apply the chain rule, we first identify the inner and outer functions. Then, we find the derivative of the inner function and multiply it by the derivative of the outer function. This gives us the derivative of the composite function.

## Can the chain rule be applied to any composite function?

Yes, the chain rule can be applied to any composite function, as long as the inner function is differentiable and the outer function is differentiable at the point where the inner function is evaluated.

## What are some common mistakes when using the chain rule?

Some common mistakes when using the chain rule include forgetting to apply the chain rule, incorrectly identifying the inner and outer functions, and making errors in the derivative of the inner or outer function. It is important to carefully follow the steps of the chain rule to avoid these mistakes.

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