Need help with combination of product and chain rule

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Homework Help Overview

The discussion revolves around differentiating functions using the product rule and chain rule, specifically focusing on the differentiation of expressions involving square roots and quotients. The original poster seeks clarification on specific steps in their differentiation process.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the application of the product rule and chain rule in differentiating the function y=\sqrt{x-2}\sqrt{x+1}. The original poster expresses confusion about a specific algebraic manipulation in their textbook's solution.
  • Another differentiation problem involving the quotient rule is introduced, with participants discussing the application of the quotient rule and the chain rule in the context of differentiating y=\frac{3-x}{\sqrt{x^{2}-2x}}.
  • Questions arise regarding the reasoning behind certain steps in the differentiation process, particularly concerning the multiplication of terms and the use of different derivative forms.

Discussion Status

The discussion is ongoing, with participants providing insights and clarifications on the differentiation rules. Some participants have expressed understanding of certain steps, while others continue to seek clarity on specific aspects of the differentiation process.

Contextual Notes

Participants are working within the constraints of homework guidelines that require the use of specific differentiation rules. There is an acknowledgment of the complexity of mastering multiple derivative rules simultaneously.

Checkfate
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I need help differentiating [tex]y=\sqrt{x-2}\sqrt{x+1}[/tex]

I am using a mix of the chain rule and product rule which my textbook for school wants me to use for this. So suggestions for different ways of approaching it won't help :P Anyways, thanks in advance for looking over it. It's my first attempt at posting in latex so let's hope it works!

For my work I have:

(dy)/(dx)=[tex](x-2)^{1/2}\frac{d}{dx}(x+1)^{1/2}+(x+1)^{1/2}\frac{d}{dx}(x-2)^{1/2}[/tex]

=[tex](x-2)^{1/2}(\frac{1}{2})(x+1)^{-1/2}+(x+1)^{1/2}(\frac{1}{2})(x-2)^{-1/2}[/tex]

And that is exactly what they have in my textbook, so that's good. But...

For the third part my textbook has

=[tex](\frac{1}{2})(x+1)^{-1/2}(x-2)^{\frac{-1}{2}}(x-2+x+1)[/tex]

This is the step that I am not understanding, it's simple algebra but I am still not quite getting it :( Can someone help?

Then the last step is

=[tex]\frac{2x-1}{2(x+1)^{1/2}(x-2)^{1/2}}[/tex] Which I somewhat understand, but not really.
\

PS - For the third step I am getting

=[tex](x-2)^{1/2}(\frac{1}{2})(x+1)^{-1/2}+(\frac{1}{2})(x+1)^{1/2}(x-2)^{-1/2}[/tex]
 
Last edited:
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For the third part, you're multiplying and dividing by [tex](x+1)^{-1/2}(x-2)^{\frac{-1}{2}}[/tex]. Try it. The fourth step is just writing the previous expression in a neat way (Remember, [tex]x^{-n} = \frac{1}{x^n}[/tex]).

Btw, is it 3 or 2 that's in the first term? ;)
 
Fixed :). Now I got to sit and think for a bit.
 
Aha! Finally I get it, simply set them up as 1/whatever^(1/2) then cross multiply to add :P Don't know why it didn't sink in immediately after I read your post but sometimes math is like that I guess. Thankyou Neutrino.

Anyways I have one more question for the day..

The Question is:

Differentiate y=[tex]\frac{3-x}{\sqrt{x^{2}-2x}}[/tex]

So here I am practicing the quotient rule instead of the product rule... I could of course use the product rule on y=(3-x)(x^2-2x)^(-1) but that would mean that I don't learn how to use the quotient rule. :P

The quotient rule is, of course [tex]\frac {v\frac {du}{dx} - u \frac {dv}{dx}}{v^{2}}[/tex]

Anyways, (this is an example problem by the way, not a homework question) For their first step they have:

(dy)/(dx)=[tex]\frac {\sqrt{x^{2}-2x} \frac {d}{dx}(3-x) - (3-x) \frac {d}{dx} (\sqrt{x^{2}-2x}}{x^{2}-2x}[/tex]

I understand this. The derivative of y with respect to x is [tex]\sqrt{x^{2}-2x}[/tex] multiplied by the derivative of (3-x) (which is of course -1) minus (3-x) multiplied by the derivative of [tex]\sqrt{x^{2}-2x}[/tex] (which is (2x-2)) But then for their second step they have:

[tex]\frac {(x^{2}-2x)^{1/2}(-1)-(3-x)(\frac{1}{2})(x^{2}-2x)^{-1/2}(2x-2)}{(x^2-2x)}[/tex]

I understand the first part on the numerator, (x^2-2x)^(1/2)(-1) since that is exactly what I stated should happen, the x^2-2x term is being multiplied by the derivative of (3-x). But then the second part I don't quite understand... Why is (3-x) being multiplied by the derivative AND another form of the derivative? The power rule states that (x^2-2x)^(1/2) has a derivative of (1/2)(x^2-2x)^(-1/2) or the nx^(n-1)... I think this is being caused by me trying to grasp all the rules of the derivative in one day, right now it's like a mish mash of semi-remembered ideas. lol.
 
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Checkfate said:
Why is (3-x) being multiplied by the derivative AND another form of the derivative? The power rule states that (x^2-2x)^(1/2) has a derivative of (1/2)(x^2-2x)^(-1/2) or the nx^(n-1)... I think this is being caused by me trying to grasp all the rules of the derivative in one day, right now it's like a mish mash of semi-remembered ideas. lol.
The chain rule.

Suppose [tex]t = \sqrt{x^2-2x}[/tex] and [tex]u = x^2 - 2x[/tex].
Therefore, [tex]t = u^\frac{1}{2}[/tex]. Now, use the chain rule to differentiate t wrt x.
 
This is perhaps best handled by the logarithm rule.
 
Thanks

Thanks a lot guys. Neutrino, much appreciated. I woke up today and read this thread and my mistake is much more obvious now. For some reason I was thinking that I was only using the power rule but of course the power rule is only for equations such as y=x^2 whereas the chain rule is for equations such as [tex]\sqrt {x-a}[/tex]. Anyways, thanks!
 

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