# Thread: Chain Rule with Three Nested Square Roots

1. ## Chain Rule with Three Nested Square Roots

This one makes my brain spin:

Find the derivative:

$\displaystyle y = \sqrt{x + \sqrt{x + \sqrt{x}}}$

I see four composed functions here (sorry I'm running out of letters at the end):
$\displaystyle y = f(u) = \sqrt{u}$
$\displaystyle u = g(x) = x + \sqrt{x}$
$\displaystyle x = h(t) = t + \sqrt{t}$
$\displaystyle t = z(v) = \sqrt {v}$

Since I probably have already made a mistake, I'll just stop here for now to see what people think of this.

Thanks.

2. Assuming that I have the functions identified correctly, the derivative calculation should work like this. This is really messy, so I'll split the derivatives out for now.

$\displaystyle y' = [f'(u)][g'(x)][h'(t)][z'(v)]$

$\displaystyle f'(u) = [\frac{1}{2}(\sqrt{x+\sqrt{x+\sqrt{x}}}^{-1/2}]$
$\displaystyle g'(x) = [1 + \frac{1}{2}x^{-1/2}]$
$\displaystyle h'(t) = [1 + \frac{1}{2}t^{-1/2}]$
$\displaystyle z'(v) = \frac{1}{2}v^{-1/2}$

Am I doing this right?

3. $\displaystyle \displaystyle y = \left[x + \left(x + x^{1/2})\right^{1/2}\right]^{1/2}$.

Let $\displaystyle \displaystyle u = x + \left(x + x^{1/2}\right)^{1/2}$ so that $\displaystyle \displaystyle y = u^{1/2}$.

$\displaystyle \displaystyle \frac{dy}{du} = \frac{1}{2}u^{-1/2} = \frac{1}{2\sqrt{x + \sqrt{x + \sqrt{x}}}}$.

$\displaystyle \displaystyle \frac{du}{dx} = 1 + \frac{d}{dx}\left[(x + x^{1/2})^{1/2}\right]$.

Another application of the Chain Rule will lead you to $\displaystyle \displaystyle \frac{du}{dx}$, which you can then use to find $\displaystyle \displaystyle \frac{dy}{dx} = \frac{du}{dx}\,\frac{dy}{du}$.

4. Thanks. I thought that I had this, but I still don't get the right answer (or so says the website). Here is what I wound up with.

$\displaystyle y' = [\frac{1}{2\sqrt{x+\sqrt{x+\sqrt{x}}}}][1 + \frac{1}{2\sqrt{x+\sqrt{x}}}][1+\frac{1}{2\sqrt{x}}]$

I must be still missing something, but at this point, I give up.

Thanks for you help, though. I really appreciate it.

5. If $\displaystyle f(x) = \sqrt{x+\sqrt{x+\sqrt{x}}}$, then using $\displaystyle [\sqrt{f(x)}]' = \frac{f'(x)}{2\sqrt{f(x)}}$ does it in three steps:

$\displaystyle \displaystyle f'(x) = \frac{\left(x+\sqrt{x+\sqrt{x}}\right)'}{2\sqrt{x+ \sqrt{x+\sqrt{x}}}} = \frac{1+\frac{\left(x+\sqrt{x}\right)'}{2\sqrt{x+\ sqrt{x}}}}{2\sqrt{x+\sqrt{x+\sqrt{x}}}} = \frac{1+\frac{1+\frac{1}{2\sqrt{x}}}{2\sqrt{x+\sqr t{x}}}}{2\sqrt{x+\sqrt{x+\sqrt{x}}}}.$

It seems you only had a bracket mistake there though (a common issue with the chain rule).

6. Originally Posted by joatmon
$\displaystyle y' = [\frac{1}{2\sqrt{x+\sqrt{x+\sqrt{x}}}}][1 + \frac{1}{2\sqrt{x+\sqrt{x}}}][1+\frac{1}{2\sqrt{x}}]$
Thanks for you help, though. I really appreciate it.
I thought I should elaborate where your bracket mistake is. It should be:

$\displaystyle \displaystyle y' = \left[\frac{1}{2\sqrt{x+\sqrt{x+\sqrt{x}}}}\right]\left[1 + \frac{1}{2\sqrt{x+\sqrt{x}}}\left(1+\frac{1}{2\sqr t{x}}\right)\right].$