derivative help

jmfine

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Sep 29, 2012
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While practicing derivative problems, I came across one that stumped me: lim h->0 f(x)=1/sqrt(x), at the point a=1/4

I got as far as setting up the equation: (1/sqrt(x+h))-(1/sqrt(x))/h. I know the first step is to take the conjugate (which I understand(; though, this is what I seem to keep messing up. I am assuming the arithmetic is what is hindering my progress. Help walking through this on this specific problem would be greatly appreciated. This is not homework, but I want to make sure there are no fraction/radical rules when doing this that I am forgetting. I would like to note that I have searched this problem elsewhere, only to find the part after the conjugate has been taken. However, that is the confusing part to me. Thank you ahead of time, for any help provided.
 
While practicing derivative problems, I came across one that stumped me: lim h->0 f(x)=1/sqrt(x), at the point a=1/4

I got as far as setting up the equation: (1/sqrt(x+h))-(1/sqrt(x))/h. I know the first step is to take the conjugate (which I understand(; though, this is what I seem to keep messing up. I am assuming the arithmetic is what is hindering my progress. Help walking through this on this specific problem would be greatly appreciated. This is not homework, but I want to make sure there are no fraction/radical rules when doing this that I am forgetting. I would like to note that I have searched this problem elsewhere, only to find the part after the conjugate has been taken. However, that is the confusing part to me. Thank you ahead of time, for any help provided.
I solved the wrong problem as SK and then mmm pointed out. Thanks to both of them for taking the trouble to figure out where I went off the rails.
 
Last edited:
While practicing derivative problems, I came across one that stumped me: lim h->0 f(x)=1/sqrt(x), at the point a=1/4

I got as far as setting up the equation: (1/sqrt(x+h))-(1/sqrt(x))/h. I know the first step is to take the conjugate (which I understand(; though, this is what I seem to keep messing up. I am assuming the arithmetic is what is hindering my progress. Help walking through this on this specific problem would be greatly appreciated. This is not homework, but I want to make sure there are no fraction/radical rules when doing this that I am forgetting. I would like to note that I have searched this problem elsewhere, only to find the part after the conjugate has been taken. However, that is the confusing part to me. Thank you ahead of time, for any help provided.

\(\displaystyle \dfrac{f(a+h) - f(a)}{h} \ = \ \dfrac{\frac{1}{\sqrt{a+h}} \ - \ \frac{1}{\sqrt{a}}}{h} \)

\(\displaystyle = \dfrac{\sqrt{a} \ - \sqrt{a+h}}{h*\sqrt{a(a+h)}} \)

Now multiply top and bottom by the "radical conjugate"

\(\displaystyle = \dfrac{\sqrt{a} \ - \sqrt{a+h}}{h*\sqrt{a(a+h)}} \ * \ \dfrac{\sqrt{a} \ + \sqrt{a+h}}{\sqrt{a} \ + \sqrt{a+h}}\)

\(\displaystyle = \dfrac{a \ - \ (a+h)}{h*\sqrt{a(a+h)}} \ * \ \dfrac{1}{\sqrt{a} \ + \sqrt{a+h}}\)

\(\displaystyle = \ - \ \dfrac{h}{h*\sqrt{a(a+h)}} \ * \ \dfrac{1}{\sqrt{a} \ + \sqrt{a+h}}\)

\(\displaystyle = \ - \ \dfrac{1}{\sqrt{a(a+h)}} \ * \ \dfrac{1}{\sqrt{a} \ + \sqrt{a+h}}\)

Now calculate the limit value by settinh "h = 0" - and - simplify.....
 
practicing derivative problems

lim h->0 f(x)=1/sqrt(x), at the point a=1/4

This problem statement is somewhat lacking.


I got as far as setting up the equation: (1/sqrt(x+h))-(1/sqrt(x))/h

Oops -- that's not an equation. Please be complete, initially.

f`(x) = lim h→0 [(1/sqrt(x+h))-(1/sqrt(x))/h]

When working with this definition, we can focus on the numerator by factoring out 1/h. This simplifies the notation.

(1/h)((1/sqrt(x+h) - 1/sqrt(x))

So the strategy is to now find a way to factor out an h from 1/sqrt(x + h) - 1/sqrt(x) because that will cancel with the factor of 1/h, and the issue of division by zero is moot. Does this make sense, thus far?


know the first step is to take the conjugate (which I understand); though, this is what I seem to keep messing up.

I am assuming the arithmetic is what is hindering my progress.

I assume that "take the conjugate" means "multiply by the conjugate". It's hard to tell where you messed up because you did not show any work. I also assume that it's not arithmetic at issue, but algebraic manipulations.

Again, I don't know what you did, but you could try it this way.

(1) Combine 1/sqrt(x+h) - 1/sqrt(x) into a single ratio

(2) Multiply top and bottom of the result by the conjugate of the numerator

(3) Simplify the resulting numerator; it should rationalize to -h

(4) Cancel factors (1/h)(-h)

(5) You should now have lim h→0 [-1/(conjugate*denom)]

(6) Substitute h = 0 and simplify

If you're interested in continued tutoring, please post your result for step (1) and set up for step (2).

Cheers :cool:
 
Ah -- I see Jeff's posting from a few minutes ago.

I think Jeff's work uses f(x) = sqrt(x) instead of 1/sqrt(x).

Anyways, his result does not agree with mine. :cool:
 
While practicing derivative problems,
I came across one that stumped me:

lim h->0 f(x)=1/sqrt(x), at the point a=1/4

Here is an alternative if you know how to use the Binomial Theorem,
as in the expansion for (a + b)^n.


The derivative can be rewritten as:


\(\displaystyle \displaystyle\lim_{h\to \ 0}\frac{1}{h}\bigg[(x + h)^{-1/2} \ - \ x^{-1/2}\bigg] \ =\)


\(\displaystyle \displaystyle\lim_{h \to \ 0}\frac{1}{h}\bigg[x^{\frac{-1}{2}} \ - \ \frac{1}{2}x^{\frac{-3}{2}}h \ + \ (the \ \ rest \ \ of \ \ terms \ \ which \ \ are \ \ at \ \ least \ \ degree \ \ 2 \ \ for \ \ h) \ \ - \ \ x^{\frac{-1}{2}}\bigg] \ = \)



\(\displaystyle \displaystyle\lim_{h \to \ 0}\frac{1}{h}\bigg[ \ - \frac{1}{2}x^{\frac{-3}{2}}h \ + \ (the \ \ rest \ \ of \ \ terms \ \ which \ \ are \ \ at \ \ least \ \ degree \ \ 2 \ \ for \ \ h) \bigg] \ = \)



\(\displaystyle \displaystyle\lim_{h \to \ 0}\bigg[ - \frac{1}{2}x^{\frac{-3}{2}} \ + \ (the \ \ rest \ \ of \ \ terms \ \ which \ \ are \ \ at \ \ least \ \ degree \ \ 1 \ \ for \ \ h) \bigg] \ = \)



\(\displaystyle - \dfrac{1}{2}x^{\frac{-3}{2}} \ = \)



\(\displaystyle \dfrac{-1}{2x^{\frac{3}{2}}} \ = \)



\(\displaystyle \dfrac{-1}{2\sqrt{x^3}}\)



You still have to substitute in 1/4 for x.
 
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