Need help with combination of product and chain rule

1. Sep 13, 2006

Checkfate

I need help differentiating $$y=\sqrt{x-2}\sqrt{x+1}$$

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 differant 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 lets hope it works!

For my work I have:

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

=$$(x-2)^{1/2}(\frac{1}{2})(x+1)^{-1/2}+(x+1)^{1/2}(\frac{1}{2})(x-2)^{-1/2}$$

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

For the third part my textbook has

=$$(\frac{1}{2})(x+1)^{-1/2}(x-2)^{\frac{-1}{2}}(x-2+x+1)$$

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

=$$\frac{2x-1}{2(x+1)^{1/2}(x-2)^{1/2}}$$ Which I somewhat understand, but not really.
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PS - For the third step I am getting

=$$(x-2)^{1/2}(\frac{1}{2})(x+1)^{-1/2}+(\frac{1}{2})(x+1)^{1/2}(x-2)^{-1/2}$$

Last edited: Sep 13, 2006
2. Sep 13, 2006

neutrino

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

Btw, is it 3 or 2 that's in the first term? ;)

3. Sep 13, 2006

Checkfate

Fixed :). Now I gotta sit and think for a bit.

4. Sep 13, 2006

Checkfate

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=$$\frac{3-x}{\sqrt{x^{2}-2x}}$$

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 $$\frac {v\frac {du}{dx} - u \frac {dv}{dx}}{v^{2}}$$

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

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

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

$$\frac {(x^{2}-2x)^{1/2}(-1)-(3-x)(\frac{1}{2})(x^{2}-2x)^{-1/2}(2x-2)}{(x^2-2x)}$$

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 derivitive 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.

Last edited: Sep 13, 2006
5. Sep 13, 2006

neutrino

The chain rule.

Suppose $$t = \sqrt{x^2-2x}$$ and $$u = x^2 - 2x$$.
Therefore, $$t = u^\frac{1}{2}$$. Now, use the chain rule to differentiate t wrt x.

6. Sep 13, 2006

arildno

This is perhaps best handled by the logarithm rule.

7. Sep 13, 2006

Checkfate

Thanks

Thanks alot 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 $$\sqrt {x-a}$$. Anyways, thanks!!