Need help with combination of product and chain rule

In summary, the conversation discusses the use of the chain rule and product rule to differentiate the function y=\sqrt{x-2}\sqrt{x+1}. The conversation also includes an example problem using the quotient rule to differentiate y=\frac{3-x}{\sqrt{x^{2}-2x}}. The use of the chain rule and logarithm rule are also mentioned in the conversation.
  • #1
Checkfate
149
0
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]
 
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  • #2
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? ;)
 
  • #3
Fixed :). Now I got to sit and think for a bit.
 
  • #4
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 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.
 
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  • #5
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.
 
  • #6
This is perhaps best handled by the logarithm rule.
 
  • #7
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!
 

1. What is the product rule and how does it work?

The product rule is a formula used in calculus to find the derivative of a product of two functions. It states that the derivative of a product of two functions is equal to the first function multiplied by the derivative of the second function, plus the second function multiplied by the derivative of the first function. In other words, (f * g)' = f'g + fg'.

2. How is the chain rule used in conjunction with the product rule?

The chain rule is used to find the derivative of composite functions, where one function is applied to the result of another function. In the case of the product rule, the chain rule is used when one or both of the functions involved are composite functions. The chain rule can be applied to each function separately, and then the product rule is used to find the derivative of the entire product.

3. Can you provide an example of using the product rule and chain rule together?

Sure, let's say we have the function f(x) = (2x + 1)^2. Using the chain rule, we can rewrite this as f(x) = u^2, where u = 2x + 1. Then, using the product rule, the derivative of f(x) would be f'(x) = 2u * u' = 2(2x + 1) * 2 = 4(2x + 1).

4. What are some common mistakes made when using the product rule and chain rule?

One common mistake is forgetting to apply the chain rule when one or both of the functions involved are composite functions. Another mistake is not distributing the derivative correctly when using the product rule.

5. In what real-world situations are the product rule and chain rule used?

The product rule and chain rule are used in many fields of science and engineering, such as physics, chemistry, and economics. They are used to find rates of change, optimize functions, and solve problems involving complex systems and relationships.

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