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 May 1st, 2016, 12:09 AM #1 Member   Joined: May 2016 From: Hell Posts: 39 Thanks: 0 How to differentiate square root with small number at the left and exponent inside? I'm not sure if this problem can be solved. I don't know the terminology for the math parts, or if this forum uses MathJax, so here is a link to one of the places I asked for help: http://math.stackexchange.com/questi...tiate-sqrt57x2 There have been several different answers, but I don't understand if any of them are right.
 May 1st, 2016, 12:39 AM #2 Math Team   Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244 The second answer is very easy to understand. What don't you get about it?
May 1st, 2016, 12:50 AM   #3
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Quote:
 Originally Posted by Azzajazz The second answer is very easy to understand. What don't you get about it?
After using the power and chain rules, I have 14x/5 times 1/"28x^8" inside the square thing, with 5 as the little number to the left.

This doesn't seem to be in any of the StackExchange answers. I don't understand the discrepancy between my answer and something like this one: calculus - How to differentiate $\sqrt[5]{7x^2}$? - Mathematics Stack Exchange

May 1st, 2016, 03:35 AM   #4
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Don't be put off by the square root signs. Remember to write out your original function in its most expanded form like the person has shown in the second reply to the stack exchange thread (as Azzajazz has already stated).

Quote:
 I have 14x/5 times 1/"28x^8" inside the square thing, with 5 as the little number to the left.
If the following doesn't make sense, i would strongly recommend revising some of your old content or to not do a calc course,

So, does this make sense: $\displaystyle \sqrt[5]{x} = x^{\frac{1}{5}}$

May 1st, 2016, 03:48 AM   #5
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 Originally Posted by Joppy So, does this make sense: $\displaystyle \sqrt[5]{x} = x^{\frac{1}{5}}$
Yes, it's converting the square to an exponent.

This is how I approached the problem:

$\displaystyle \sqrt[5]{7x^2}$ = (7x^2)^1/5

= [product rule] (1/5)((7x^2)^1/5 -1) = [chain rule] (1/5)(7x^2)^-4/5(14x) = (14x/5)($\displaystyle \sqrt[5]{28x^8}$)

I've tried to edit this many, many times, but the Mathjax must be glitching. Can you click the edit button to read the post?

Last edited by Incomprehensible; May 1st, 2016 at 03:53 AM.

 May 1st, 2016, 04:09 AM #6 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,764 Thanks: 623 Math Focus: Yet to find out. Before trying to apply any rules of differentiation, express your function in its most expanded form. So you got the first step right, $\displaystyle \sqrt[5]{7x^2} = (7x^2)^{\frac{1}{5}}$ Now, recall the exponent laws, $\displaystyle (a^b)^c = a^{bc}$ Now we have, $\displaystyle (7x)^{\frac{2}{5}}$ If i go any further i'll just be duplicating Andres Mejia post on math stack exchange calculus - How to differentiate $\sqrt[5]{7x^2}$? - Mathematics Stack Exchange Thanks from Incomprehensible
May 1st, 2016, 04:33 AM   #7
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Quote:
 Originally Posted by Joppy Before trying to apply any rules of differentiation, express your function in its most expanded form. So you got the first step right, $\displaystyle \sqrt[5]{7x^2} = (7x^2)^{\frac{1}{5}}$ Now, recall the exponent laws, $\displaystyle (a^b)^c = a^{bc}$ Now we have, $\displaystyle (7x)^{\frac{2}{5}}$ If i go any further i'll just be duplicating Andres Mejia post on math stack exchange calculus - How to differentiate $\sqrt[5]{7x^2}$? - Mathematics Stack Exchange
Part of the reason I'm confused is because I looked at my notes, at a similar problem, while trying to solve this one. The book I'm using is incomprehensible, so I was trying to scrape together solutions until I could match the answers given. I'm not sure if what I did in my notes here is correct or not: Imgur: The most awesome images on the Internet.

I don't understand why the answer I worked out matched the book's answer for that problem, yet I cannot solve this similar problem.

 May 1st, 2016, 04:37 AM #8 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,764 Thanks: 623 Math Focus: Yet to find out. This isn't a case of "find a problem that looks the same and use the same method", and most of the time, it never is. By all means, look at your working for questions you've solved in the past as revision and guidance, but don't just match the patterns. I suggest you look at some of your previous working, such as the image you posted just now, and for every line of your working, write out any rules/laws of algebra, differentiation you applied to the problem. Hopefully then you can start to see how it all works. Thanks from Incomprehensible
May 1st, 2016, 04:41 AM   #9
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Quote:
 Originally Posted by Incomprehensible $\displaystyle \sqrt[5]{7x^2}$ = (7x^2)^1/5 = [product rule] (1/5)((7x^2)^1/5 -1) = [chain rule] (1/5)(7x^2)^-4/5(14x) = (14x/5)($\displaystyle \sqrt[5]{28x^8}$)
A few things. Firstly, all the equalities are wrong. Those expressions aren't actually equal to each other.

Secondly, this is overly complicated and I wouldn't recommend this method on questions such as this.

That's probably why the Maths Exchange answers are different.

I agree whole-heartedly with Joppy. The beauty of maths comes from finding simple and elegant solutions, not just drilling out the same technique over and over again.

 May 1st, 2016, 04:54 AM #10 Member   Joined: May 2016 From: Hell Posts: 39 Thanks: 0 Okay, I think I see where my answer splits from the given ones. http://imgur.com/IwGW2OY I have the first part of that in my notebook, but I don't know how to properly phrase that in MathJax. The second part, I'm having trouble understanding. How would I get from the first part to the second?

 Tags differentiate, exponent, inside, left, number, root, small, square

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