Need help with combination of product and chain rule

[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 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)=(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 that's 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}

 

[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? ;)

 

[Checkfate]

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

 

[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.

 

[neutrino]

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 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.

 

[arildno]

This is perhaps best handled by the logarithm rule.

 

[Checkfate]

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