1. ## Finding the derivative

I need to find the derivative of the following function.

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

[tex] i think this can be simplified to $sqrt{x} + (sqrt{x}(sqrt{x}+1))^1/4$

so would it be reasonable to use the product and chain rule to solve the derivative of the 2nd part of the equation and just generally derive the 1st part?

2. ## Re: Finding the derivative

Do you mean to write:

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

3. ## Re: Finding the derivative

Yes! thank you!

4. ## Re: Finding the derivative

I might square the equation to get:

$y^2=x+\left(x+x^{\frac{1}{2}} \right)^{\frac{1}{2}}$

Now use implicit differentiation.

5. ## Re: Finding the derivative

Implicit differentiation works, but I don't think that it's any extra work to evaluate the derivative explicitly. Your function is \displaystyle \begin{align*} y = \sqrt{x + \sqrt{x + \sqrt{x}}} = \left[ x + \left( x + x^{\frac{1}{2}} \right)^{\frac{1}{2}} \right] ^{\frac{1}{2}} \end{align*}.

Start by letting \displaystyle \begin{align*} u = x + \left( x + x^{\frac{1}{2}} \right)^{\frac{1}{2}} \end{align*} so that \displaystyle \begin{align*} y = u^{\frac{1}{2}} \end{align*}. Then \displaystyle \begin{align*} \frac{dy}{du} = \frac{1}{2}u^{-\frac{1}{2}} = \frac{1}{2\sqrt{u}}= \frac{1}{2\sqrt{x + \sqrt{x + \sqrt{x}}}} \end{align*}.

Now as for \displaystyle \begin{align*} \frac{du}{dx} = 1 + \frac{d}{dx} \left[ \left( x + x^{\frac{1}{2}} \right)^{\frac{1}{2}} \right] \end{align*} you will need to use the chain rule again. You should get \displaystyle \begin{align*} \frac{du}{dx} = 1 + \frac{1}{2}\left( x + x^{\frac{1}{2}} \right)^{-\frac{1}{2}} \left( 1 + \frac{1}{2}x^{-\frac{1}{2}} \right) = 1 + \frac{1}{2 \sqrt{ x + \sqrt{x} } }\left( 1 + \frac{1}{ 2\sqrt{x} } \right) \end{align*}.

Therefore \displaystyle \begin{align*} \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{2\sqrt{ x + \sqrt{ x + \sqrt{x} } }} \left[ 1 + \frac{1}{2\sqrt{x + \sqrt{x}}}\left( 1 + \frac{1}{2\sqrt{x}} \right) \right] \end{align*}