Getting this derivative: f(x) = x + sqrt(x)

[MarkSA]

Hi folks,

I feel i'm understanding what we're doing in Calc 1 fairly well, but i've run into a troublesome problem. I think this is more of a simple algebra issue than anything, but I can't seem to find it.

Find the derivative of the function using the definition of derivative.

f(x) = x + sqrt(x)

Not sure how to use the proper notation on here, but basically I used the formula to find the limit as h approaches 0:
f(x+h) - f(x)/h

This gets me (x + h) + sqrt(x+h) - x - sqrt(x), all over h. Typically, the goal with this problem tends to be to cancel all terms out that don't have an h, then to factor an h out of the numerator to cancel the one in the denominator, and then substitute with 0. Well, the x's will cancel, but i'm still left with h + sqrt(x+h) - sqrt(x), all over h. I need to factor an h out of the numerator. How can I solve this problem in this fashion?

(By the way, i've heard there's possibly an easier way to find derivatives than this, but the above is what we're currently learning in the book and the teacher is strict about not using prior knowledge on homework.)

 

[tkhunny]

Re: Getting this derivative..

f(x+h) - f(x)/h

First, check your notation. Add brackets or parentheses in the numerator. It does NOT mean what you want without them.

Typically, the goal with this problem tends to be to cancel all terms out that don't have an h, then to factor an h out of the numerator to cancel the one in the denominator, and then substitute with 0.

No, no, and no. Do not confuse an early method with the goal of the process. Establishing the limit is not a trivial thought. Do NOT substitute the limit. Don't do it.

h + sqrt(x+h) - sqrt(x), all over h.

This demonstrates the problem with learning a method rather than understanding the concept. Finding the limit calls on the student to use creativity and ALL the background material that should have been learned before this point. You must think it up.

In this case, try splitting it up a bit. \(\displaystyle \frac{h}{h} + \frac{\sqrt{x+h}+\sqrt{x}}{h}\)
Now what?

 

[Deleted member 4993]

First find the derivative of \(\displaystyle sqrt{x}\)

Follow the same technique - i.e. multiplying the conjugate at the top and the bottom - combine it with the technique tkhunny described - you are home-free.

 

[MarkSA]

First, check your notation. Add brackets or parentheses in the numerator. It does NOT mean what you want without them.

[f(x+h) - f(x)]/h

No, no, and no. Do not confuse an early method with the goal of the process. Establishing the limit is not a trivial thought. Do NOT substitute the limit. Don't do it.

This isn't actually an early method in this book though. This is how we're doing it currently and it's being taught. The homework questions i'm studying are for a particular section and this is how all of the examples are worked out in this section.

First find the derivative of sqrt(x)

I believe this is the method (breaking it up and solving pieces for derivatives) that they are teaching in the next section, but I can't use it to solve this problem because of the rules. This is odd I know, but it's how they've decided to teach it (no knowledge prior or reading ahead can be used.)

I tried breaking it up, but i'm still not sure where to go from here if I want to solve this problem the way i'm supposed to in this section. Can you explain a little more?

 

[MarkSA]

Thought I had it.. didn't do it correctly. Still lost on this one. Can you elaborate more on what process I should use?

 

[Deleted member 4993]

First finding the derivative of \(\displaystyle sqrt{x}\)

\(\displaystyle \frac{\sqrt{x+h}-\sqrt{x}}{h}


= \frac{\sqrt{x+h}-\sqrt{x}}{h}\, \times \,\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}

= \frac{{x+h}-{x}}{h}\, \times \, \frac{1}{\sqrt{x+h}+\sqrt{x}}\)

Now if you take the limit - you'll find the derivative of \(\displaystyle sqrt{x}\)

Use the same technique on the expression that tkhunny found for you.

Show us your work - so that we can understand better where you are stuck.

 

[MarkSA]

Ok thanks.

So the derivative of the entire function is just 1/[2*sqrt(h)] , since deriv. of x (constant) is 0, and the sum of the derivatives is 0 + 1/[2*sqrt(h)] = 1/[2*sqrt(h)]?

If that's true, then this is using the sum rule correct? We haven't gotten to that yet in the book, so I can't use that method to solve this problem unfortunately. Is there another way?

Let me paste an example problem from my book. This is how i'm supposed to solve this problem in this section of the book (using this exact method, no other short cuts since we haven't gotten to them yet.) Do you know how I can go about finding the derivative of the troublesome problem in this way? Where i'm running into trouble is factoring an h out of the numerator, which is: h+sqrt(x+h)-sqrt(x)

If f(x) = sqrt(x - 1), find the derivative of f.
f'(x) = lim as h approaches 0 of [f(x + h) - f(x)]/h
= lim as h approaches 0 of sqrt(x + h - 1) - sqrt(x - 1)
= lim as h approaches 0 of sqrt(x + h - 1) - sqrt(x - 1) multiplied by conjugate
= lim as h approaches 0 of [(x + h - 1) - (x - 1)]/h[sqrt(x + h - 1) + sqrt(x - 1)]
= lim as h approaches 0 of 1/[sqrt(x + h - 1) + sqrt(x - 1)]
= 1/[sqrt(x - 1) + sqrt(x - 1)]
= 1/[2*sqrt(x - 1)] as the derivative

 

[Deleted member 4993]

Ok thanks.

So the derivative of the entire function is just 1/[2*sqrt(h)] , ..NO

since deriv. of x (constant - no it is not a constant) is 1,

and the sum of the derivatives is

1 + 1/[2*sqrt(x)] ?

If that's true, then this is using the sum rule correct? We haven't gotten to that yet in the book, so I can't use that method to solve this problem unfortunately. Is there another way?

Let me paste an example problem from my book. This is how i'm supposed to solve this problem in this section of the book (using this exact method, no other short cuts since we haven't gotten to them yet.) Do you know how I can go about finding the derivative of the troublesome problem in this way? Where i'm running into trouble is factoring an h out of the numerator, which is: h+sqrt(x+h)-sqrt(x)

If f(x) = sqrt(x - 1), find the derivative of f.
f'(x) = lim as h approaches 0 of [f(x + h) - f(x)]/h
= lim as h approaches 0 of sqrt(x + h - 1) - sqrt(x - 1)
= lim as h approaches 0 of sqrt(x + h - 1) - sqrt(x - 1) multiplied by conjugate
= lim as h approaches 0 of [(x + h - 1) - (x - 1)]/h[sqrt(x + h - 1) + sqrt(x - 1)]
= lim as h approaches 0 of 1/[sqrt(x + h - 1) + sqrt(x - 1)]
= 1/[sqrt(x - 1) + sqrt(x - 1)]
= 1/[2*sqrt(x - 1)] as the derivative

 

[mark]

here is a useful link that may help

http://archives.math.utk.edu/visual.cal ... /limits.7/

 

[MarkSA]

y = x with a slope of 1.. I should have seen that. Anyway, using the formula, plugging in x directly gets

[(x+h)-x]/h = h/h = 1, derivative of 1. Not sure what I was thinking.

Back to the main problem though... Subhotosh, your solution was using the sum rule, yes? As I mentioned, i'm not allowed to use that to do the homework.

We had a test last week actually where the teacher explicitly said that if we used the derivative tricks to solve for the slope of a tangent to the curve, they would be counted wrong. It's a similar case for the homework.

Can this problem be solved using the method I quoted in the post above without using future topics such as the sum rule? Is it possible, and where would I start?

Please realize that the way this book is ordered is that we are introduced to the long, inefficient, and sometimes occassionally useless methods of finding problems like limits or derivatives, and only LATER are we shown the easy ways to do these. Our homework has to reflect that. And i'm probably spending way too much time on this one since we are assigned about 50 homework problems a week and only 2 of them are graded. But i'd really like to know how to do this the way that they are asking.

 

[pka]

\(\displaystyle \L \begin{array}{rcl}
f(x) & = & x + \sqrt x \\
\frac{{f(x + h) - f(x)}}{h} & = & \frac{{\left[ {\left( {x + h} \right) + \sqrt {x + h} } \right] - \left[ {x + \sqrt x } \right]}}{h} \\
& = & \frac{{h + \left[ {\sqrt {x + h} - \sqrt x } \right]}}{h} \\
& = & \frac{{h + \left[ {\sqrt {x + h} - \sqrt x } \right]}}{h}\frac{{\left[ {\sqrt {x + h} + \sqrt x } \right]}}{{\left[ {\sqrt {x + h} + \sqrt x } \right]}} \\
& = & \frac{{h\left[ {\sqrt {x + h} + \sqrt x } \right] + \left[ h \right]}}{{h\left[ {\sqrt {x + h} + \sqrt x } \right]}} \\
& = & \frac{{\left[ {\sqrt {x + h} + \sqrt x } \right] + \left[ 1 \right]}}{{\left[ {\sqrt {x + h} + \sqrt x } \right]}} \\
\end{array}\)

\(\displaystyle \L \lim _{h \to 0} \frac{{\left[ {\sqrt {x + h} + \sqrt x } \right] + \left[ 1 \right]}}{{\left[ {\sqrt {x + h} + \sqrt x } \right]}} = \frac{{2\sqrt x + 1}}{{2\sqrt x }} = 1 + \frac{1}{{2\sqrt x }}\)

 

[stapel]

Since we don't know what your book is calling "the sum rule", I'm afraid we cannot comment on that. But I am curious as to how far you got, what you did, and where you ground to a halt with respect to this set-up for sqrt[x]:

First finding the derivative of \(\displaystyle sqrt{x}\)

. . .\(\displaystyle \L \frac{\sqrt{x\,+\,h}\,-\, \sqrt{x}}{h}\)

. . . . .\(\displaystyle \L = \, \frac{\sqrt{x\,+\,h}\,-\,\sqrt{x}}{h}\, \times \,\frac{\sqrt{x\,+\,h}\,+\,\sqrt{x}}{\sqrt{x\,+\,h}\,+\,\sqrt{x}}\)

. . . . .\(\displaystyle \L =\, \frac{{x\,+\,h}\,-\,{x}}{h}\, \times \, \frac{1}{\sqrt{x\,+\,h}\,+\,\sqrt{x}}\)

Now if you take the limit - you'll find the derivative of \(\displaystyle sqrt{x}\)...

Finding your limit would involve only adding an "(x + h) - (x)" to the beginning of what you were given above.

Please reply with specifics, showing all of your steps. Thank you!

Eliz.

 

[MarkSA]

pka: Thank you! That's exactly what I was looking for! And now I see why I couldn't figure out how to do this one, because you used some math with the conjugate in a way that i've never actually seen used before. Here I was, thinking multiplying by the conjugate would only do the trick in simple problems like:

All conjugate problems i've ever seen solved in college algebra have been of the form of, say:
[(sqrt(3x) - 4] multiplied by [(sqrt(3x) + 4] (etc)

I've never done or seen one done with 3 terms. Actually, I don't recall seeing much of a formal definition of the conjugate ever while taking, aside from the procedure to get it on simple problems like the one above.

For example, suppose I had: [h + 3x + 5x^2 + sqrt(3x) - sqrt(x+1)] as the numerator.
So, simply multiplying the numerator and denominator by [sqrt(3x) + sqrt(x+1)] as the conjugate while ignoring all of the other terms would also work algebraically?

Thanks for helping me figure out this algebra problem. :wink:

(BTW: Sorry about quoting it just as the "sum rule", I thought it would be something well known like the IVT from the way the book stated it. From what I understand (i'm merely reading ahead on it), it's saying that the sum rule means the derivative of the sum of two functions is the same as the derivatives of each function added together)

And is there an easy way to do that math script you guys are using? It would be much easier and clearer for me to write it that way if it's not too difficult to do.

 

[pka]

And is there an easy way to do that math script you guys are using? It would be much easier and clearer for me to write it that way if it's not too difficult to do.

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