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

  • #1

Homework Statement



Define f(x)=[itex]\sqrt{}(x + (\sqrt{}(x + \sqrt{}x)[/itex]
Determine where f is differentiable and compute the derivative

Homework Equations



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

The Attempt at a Solution


By the definition, f(x) = [itex]\sqrt{}x[/itex] does not have a derivative at 0, since limit as x approaches zero, x>0, is 1/[itex]\sqrt{}x[/itex], 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.
 

Answers and Replies

  • #2
280
0
Chain rule can be thought about as fractions:
Let
[tex] f(x) = \sqrt{x+\sqrt{x+\sqrt{x}}}[/tex]
and we want
[tex] \frac{df}{dx} = f'(x)[/tex]
then let
[tex] g(x) = x + \sqrt{x+\sqrt{x}}[/tex]

so

[tex] f(g(x)) = \sqrt{g(x)}[/tex]

then
[tex] \frac{df}{dg}\frac{dg}{dx} = \frac{df}{dx} = g'(x)\frac{1}{2}g(x)^{-\frac{1}{2}}[/tex]

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

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.
 
  • #3
I think the limit as x -> xo, if you multiply numerator and denominator by it conjugate you get, f'(xo) = 1/2(xo+(xo + (xo)^(1/2))^(1/2))^(1/2))

Is that it?
 
  • #4
280
0
I think the limit as x -> xo, if you multiply numerator and denominator by it conjugate you get, f'(xo) = 1/2(xo+(xo + (xo)^(1/2))^(1/2))^(1/2))

Is that it?

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

Let
[tex] f(x) = \sqrt{x+\sqrt{x+\sqrt{x}}}[/tex]

Then we seek

[tex] 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}[/tex]

but let
[tex] g(x) = x + \sqrt{x+\sqrt{x}}[/tex]

Then

[tex]Z = \frac{\sqrt{g(x)} - \sqrt{g(x_0)}}{g(x)-g(x_0)}\frac{g(x)-g(x_0)}{x-x_0}[/tex]

But let
[tex] h(x) = \sqrt{x+\sqrt{x}}[/tex]

then

[tex]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 )[/tex]

But let

[tex] y(x) = x + \sqrt{x}[/tex]

then

[tex]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 )[/tex]
and the final one falls into place:

[tex]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 )[/tex]

Taken the limit of Z as x approaches x_0. Do so for each piece.
 
  • #5
280
0
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.
 
  • #6
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.
 

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