Finding the Derivative of y=sqrt(x+sqrt(x+sqrt(x)))

[ScienceMan]

Homework Statement

This is a chain rule problem that I can't seem to get right no matter what I do. It wants me to find the derivative of y=sqrt(x+sqrt(x+sqrt(x)))

Homework Equations

dy/dx=(dy/du)*(du/dx)
d/dx sqrtx=1/(2sqrtx)
d/dx x=1
(f(x)+g(x))'=f'(x)+g'(x)

The Attempt at a Solution

My attempted solution is dy/dx=1/(2sqrt(x+sqrt(x+sqrtx)))*(1+(1/(2sqrt(x+sqrtx))))*(1+(1/(2sqrtx))).

I took the derivative of the outermost function and left the inner function alone, then multiplied by the derivative of the inner function, and continuing until I reached the innermost function. I'm doing exactly what I was told to do in class and this approach has worked on all the other problems so far, but my online homework system isn't taking this as a solution. Where am I going wrong?

 

[scottdave]

What result did you get (what are submitting for an answer). Can you outline your steps? Have you checked against something like Wolframalpha?

 

[ScienceMan]

What result did you get (what are submitting for an answer). Can you outline your steps? Have you checked against something like Wolframalpha?

I'm submitting exactly what I put in for the answer I got. I know formatting isn't the problem, since it would tell me if it was (I'm using WebWorks in case that matters.) I did check it against Wolfram Alpha, and you can see the result here. As far as I can tell the results are the same, but I might be wrong on that as WA is giving the answer in an odd format instead of showing the factors being multiplied side by side.

My steps were:

1. Take the derivative of y with respect to u of the outermost square root, with u being everything inside that, giving me 1/(2sqrt(x+sqrt(x+sqrtx)))
2. Take the derivative of u with respect to x, with a new u being anything inside the 2nd most outer square root, giving me (1+(1/(2sqrt(x+sqrtx))))
3. Take the derivative of the new u with respect to x, giving me (1+(1/(2sqrtx)))
4. Multiply them all together.

 

[scottdave]

I think the step 4. multiply all together, may be where you might have issues. It's a little vague. What answer did you arrive at?
Try working from the inside out:

We have y=sqrt(x+sqrt(x+sqrt(x)))
set u = x+sqrt(x), so du/dx = (1 + 1/(2sqrt(x)))
set v = x + sqrt(u), so dv/dx = { 1 + 1/(2sqrt(u)) }*(du/dx)
Now y = sqrt(v). so dy/dx = 1/(2sqrt(v)) * (dv/dx)
Substitute in for u, v and du/dx and dv/dx.

You could try picking different values for x, and then estimating the slope by calculating Δy / Δx, picking sufficiently small values for Δx.
See how these values compare with your expression for dy/dx.

If all of that agrees, and the automated homework system still won't accept it, then it's time to contact your instructor and see if they provide guidance.

 

[scottdave]

Here are some values that I got for Δy / Δx, using Δx = 1 x 10^-5:
x = 1, Δy / Δx = 0.4925
x = 2, Δy / Δx = 0.3483
x = 3, Δy / Δx = 0.2849
x = 4, Δy / Δx = 0.2471
x= 10, Δy / Δx = 0.1571

These are plugging into the your original expression for y, not even doing the derivative. If your expression for derivative is correct, then you should be able to plug in x values and get close to these numbers for dy/dx.

 

[Ray Vickson]

Homework Statement

This is a chain rule problem that I can't seem to get right no matter what I do. It wants me to find the derivative of y=sqrt(x+sqrt(x+sqrt(x)))

Homework Equations

dy/dx=(dy/du)*(du/dx)
d/dx sqrtx=1/(2sqrtx)
d/dx x=1
(f(x)+g(x))'=f'(x)+g'(x)

The Attempt at a Solution

My attempted solution is dy/dx=1/(2sqrt(x+sqrt(x+sqrtx)))*(1+(1/(2sqrt(x+sqrtx))))*(1+(1/(2sqrtx))).

I took the derivative of the outermost function and left the inner function alone, then multiplied by the derivative of the inner function, and continuing until I reached the innermost function. I'm doing exactly what I was told to do in class and this approach has worked on all the other problems so far, but my online homework system isn't taking this as a solution. Where am I going wrong?

You are doing nothing wrong; if the answer you submitted is exactly what you have typed here then it should be acceptable, unless there are some unbalanced parentheses----always a possibility when you have so many nested together like you do. As others have suggested, maybe there is also a formatting issue.

 

[scottdave]

Maybe I missed your typed out solution before? I was on my phone when I originally saw your post. I looked at your expression. I am not sure if your expression is right though.

 

[ScienceMan]

I figure I should post the end result of this. I ended up asking my instructor and the problem was I needed to move some parentheses around. I guess I messed those up somehow and it ended up screwing up the answer somehow. Thanks for the help everyone.