Find derivative of Square root (x + square root(x + x^(1/2))) Help

[IntroAnalysis]

Homework Statement

Define f(x)=$\sqrt{}(x + (\sqrt{}(x + \sqrt{}x)$
Determine where f is differentiable and compute the derivative

Homework Equations

f'(x_o)= lim as x approaches x_o (f(x) - f(x_o))/(x - x_o)

The Attempt at a Solution

By the definition, f(x) = $\sqrt{}x$ does not have a derivative at 0, since limit as x approaches zero, x>0, is 1/$\sqrt{}x$, so can't have zero in denominator, not differentiable at 0.

I think above function, because of Algebra of Derivatives would also not be differentiable at zero. I keep ending up with a big mess trying to find it's derivative, I think I'm missing a rule somewhere, any suggestions are welcome.

 

[RoshanBBQ]

Chain rule can be thought about as fractions:
Let
$$f(x) = \sqrt{x+\sqrt{x+\sqrt{x}}}$$
and we want
$$\frac{df}{dx} = f'(x)$$
then let
$$g(x) = x + \sqrt{x+\sqrt{x}}$$

so

$$f(g(x)) = \sqrt{g(x)}$$

then
$$\frac{df}{dg}\frac{dg}{dx} = \frac{df}{dx} = g'(x)\frac{1}{2}g(x)^{-\frac{1}{2}}$$

Let me start you off on the next step and see if you can finish it. Let
$$h(x) = \sqrt{x+\sqrt{x}}$$
then
$$g(x) = x + h(x)$$
$$\frac{dg}{dx} = 1 + \frac{dh}{dx}$$

Hint: use the same logic already presented to find the derivative of h(x). Then substitute all of your answers into previous equations for the final answer.

 

[IntroAnalysis]

I think the limit as x -> x_o, if you multiply numerator and denominator by it conjugate you get, f'(x_o) = 1/2(x_o+(x_o + (x_o)^(1/2))^(1/2))^(1/2))

Is that it?

 

[RoshanBBQ]

I think the limit as x -> x_o, if you multiply numerator and denominator by it conjugate you get, f'(x_o) = 1/2(x_o+(x_o + (x_o)^(1/2))^(1/2))^(1/2))

Is that it?

I apologize. I didn't see you were dealing with the limit.

Let
$$f(x) = \sqrt{x+\sqrt{x+\sqrt{x}}}$$

Then we seek

$$Z = \frac{f(x)-f(x_0)}{x-x_0}=\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x_0+\sqrt{x_0+\sqrt{x_0}}}}{x-x_0}$$

but let
$$g(x) = x + \sqrt{x+\sqrt{x}}$$

Then

$$Z = \frac{\sqrt{g(x)} - \sqrt{g(x_0)}}{g(x)-g(x_0)}\frac{g(x)-g(x_0)}{x-x_0}$$

But let
$$h(x) = \sqrt{x+\sqrt{x}}$$

then

$$Z = \frac{\sqrt{g(x)} - \sqrt{g(x_0)}}{g(x)-g(x_0)}\frac{x+h(x)-(x_0+h(x_0))}{x-x_0}=\frac{\sqrt{g(x)} - \sqrt{g(x_0)}}{g(x)-g(x_0)} \left ( \frac{x-x_0}{x-x_0} + \frac{h(x) - h(x_0)}{x-x_0} \right )$$

But let

$$y(x) = x + \sqrt{x}$$

then

$$Z =\frac{\sqrt{g(x)} - \sqrt{g(x_0)}}{g(x)-g(x_0)} \left ( \frac{x-x_0}{x-x_0} + \frac{\sqrt{y(x)} - \sqrt{y(x_0)}}{y(x)-y(x_0)}\frac{y(x)-y(x_0)}{x-x_0} \right )$$
and the final one falls into place:

$$Z =\frac{\sqrt{g(x)} - \sqrt{g(x_0)}}{g(x)-g(x_0)} \left ( \frac{x-x_0}{x-x_0} + \frac{\sqrt{y(x)} - \sqrt{y(x_0)}}{y(x)-y(x_0)}\left ( \frac{x-x_0}{x-x_0} +\frac{\sqrt{x}-\sqrt{x_0}}{x-x_0} \right ) \right )$$

Taken the limit of Z as x approaches x_0. Do so for each piece.

 

[RoshanBBQ]

If you don't know how to handle the squareroot case frequent in that final equation I wrote, consider x - x_0 as a difference of squares.

 

[IntroAnalysis]

If you don't know how to handle the squareroot case frequent in that final equation I wrote, consider x - x_0 as a difference of squares.

Thank you. Your approach makes sense and is clear.