Can someone help me find the derivative of y=$\sqrt{x+\sqrt{x+\sqrt{x}}}$?

[lastochka]

Hi, this is my online homework exercise. I think I did it right, but the system gives me wrong answer...
Can someone please check what I did
y=$\sqrt{x+\sqrt{x+\sqrt{x}}}$

y$^{\prime}$=$\frac{1}{2}$((x+(x+${x}^{\frac{1}{2}}$)$^{\frac{1}{2}}$)$^\frac{-1}{2}$(1+$\frac{1}{2}$(x+${x}^{\frac{1}{2}}$))$^\frac{-1}{2}$(1+$\frac{1}{2}$x$^\frac{-1}{2}$))
Thank you in advance!

 

[Dethrone]

Is this your answer?

View attachment 4049

If no, then we can work towards getting that answer. :D

 

[lastochka]

Yes, this is what I have so far...
I am not sure how to do it your way.
Is what I did completely wrong??

 

[I like Serena]

Hi, this is my online homework exercise. I think I did it right, but the system gives me wrong answer...
Can someone please check what I did
y=$\sqrt{x+\sqrt{x+\sqrt{x}}}$

y$^{\prime}$=$\frac{1}{2}$((x+(x+${x}^{\frac{1}{2}}$)$^{\frac{1}{2}}$)$^\frac{-1}{2}$(1+$\frac{1}{2}$(x+${x}^{\frac{1}{2}}$))$^\frac{-1}{2}$(1+$\frac{1}{2}$x$^\frac{-1}{2}$))
Thank you in advance!

Hi lastochka,

You seem to have a parenthesis in the wrong place.
It should be:
y$^{\prime}$=$\frac{1}{2}$(x+(x+${x}^{\frac{1}{2}}$)$^{\frac{1}{2}}$)$^\frac{-1}{2}$(1+$\frac{1}{2}$(x+${x}^{\frac{1}{2}}$)$^\frac{-1}{2}$(1+$\frac{1}{2}$x$^\frac{-1}{2}$))

 

[Dethrone]

$$\d{}{x}\sqrt{x+\sqrt{x+\sqrt{x}}}$$

Let $g(x)=\sqrt{x+\sqrt{x}}$ (for ease of reading)

Now, we have:
$$\d{}{x}\sqrt{x+g(x)}$$
$$=\frac{1}{2\sqrt{x+g(x)}}\cdot (1+g'(x))$$Evaluate $g'(x)$
$$g'(x)=\frac{1}{2\sqrt{x+\sqrt{x}}}\cdot (1+\frac{1}{2\sqrt{x}})=\frac{1+\frac{1}{2\sqrt{x}}}{2\sqrt{x+\sqrt{x}}}$$

Combining the answers:

$$\d{}{x}\sqrt{x+g(x)}=\frac{ (1+g'(x))}{2\sqrt{x+g(x)}}=\frac{ 1+\frac{1+\frac{1}{2\sqrt{x}}}{2\sqrt{x+\sqrt{x}}}}{2\sqrt{x+\sqrt{x+\sqrt{x}}}}$$

 

[Prove It]

Hi, this is my online homework exercise. I think I did it right, but the system gives me wrong answer...
Can someone please check what I did
y=$\sqrt{x+\sqrt{x+\sqrt{x}}}$

y$^{\prime}$=$\frac{1}{2}$((x+(x+${x}^{\frac{1}{2}}$)$^{\frac{1}{2}}$)$^\frac{-1}{2}$(1+$\frac{1}{2}$(x+${x}^{\frac{1}{2}}$))$^\frac{-1}{2}$(1+$\frac{1}{2}$x$^\frac{-1}{2}$))
Thank you in advance!

Implicit differentiation may be easiest to use in this case:

$\displaystyle \begin{align*} y &= \sqrt{ x + \sqrt{ x + \sqrt{ x } } } \\ y^2 &= x + \sqrt{ x + \sqrt{x}} \\ y^2 - x &= \sqrt{ x + \sqrt{x}} \\ \left( y^2 - x \right) ^2 &= x + \sqrt{x} \\ y^4 - 2x\,y^2 + x^2 &= x + \sqrt{x} \\ \frac{\mathrm{d}}{\mathrm{d}x} \, \left( y^4 - 2x\,y^2 + x^2 \right) &= \frac{\mathrm{d}}{\mathrm{d}x} \, \left( x + \sqrt{x} \right) \\ 4y^3\,\frac{\mathrm{d}y}{\mathrm{d}x} - 2 \left( y^2 + 2x\,y\,\frac{\mathrm{d}y}{\mathrm{d}x} \right) + 2x &= 1 + \frac{1}{2\,\sqrt{x}} \\ 4y^3\,\frac{\mathrm{d}y}{\mathrm{d}x} - 2y^2 - 4x\,y\,\frac{\mathrm{d}y}{\mathrm{d}x} + 2x &= 1 + \frac{1}{2\,\sqrt{x}} \\ 4y \, \left( y^2 - x \right) \, \frac{\mathrm{d}y}{\mathrm{d}x} &= 1 + \frac{1}{2\,\sqrt{x}} - 2x + 2y^2 \\ \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{1 + \frac{1}{2\,\sqrt{x}} - 2x +2y^2}{4y\,\left( y^2 - x \right) } \end{align*}$

and when you substitute y in and simplify you should get the same as the answer given.

 

[lastochka]

Thank you, everyone for your help!