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Complex Derivative - Easier Method?

  • Thread starter SuspectX
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Homework Statement


Find the derivative of:

[itex]\frac{(x^{3} + 5x^{2} − 9x + 4)^{3/4}}{(x^{4} − x^{3} + x^{2} − x + 1)([x^{5} − x^{-5}]^{1/2})}[/itex]

Homework Equations


Product Rule
Quotient Rule

The Attempt at a Solution


I've attempted using the quotient rule for the entire thing and gotten a big mess. Is there an easier way which may include breaking it down first?

So basically I wanted to use the quotient rule, meaning I need the derivative of the numerator and denominator and I got...

Numerator:
[itex]\frac{3/4{(3x^{2} + 10x - 9)(x^{3} + 5x^{2} - 9x + 4)^{-1/4}}}{}[/itex]


Denominator:
[itex]\frac{1/2(5x^{4} + 5x^{-6})(x^{5} - x^{-5})^{-1/2}(x^{4}-x^{3}+x^{2}-x+1) + ([x^{5} − x^{-5}]^{1/2})(4x^{3} - 3x^{2} + 2x - 1)}{}[/itex]
 
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Answers and Replies

  • #2
SammyS
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Show us your attempt.

It helps us to help you.

Added in Edit:

Any way you do this will be a big mess.

It may help to find the derivative of the denominator separately (It will require use of the product rule.), then simplify it and use that result in the overall derivative.
 
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  • #3
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I've edited my original post to show what I have so far. I can't seem to find a way or simplifying it, so I'm kinda worried I did something wrong...
 
  • #4
SammyS
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The only thing that I have come up with is the following:

[itex]\displaystyle x^5-x^{-5}=\frac{x^{10}-1}{x^5}[/itex]

[itex]x^{10}-1=(x^2-1)(x^4+x^3+x^2+x+1)(x^4-x^3+x^2-x+1)[/itex]
[itex]=(x^6+x^5-x-1)(x^4-x^3+x^2-x+1)[/itex]​

So, you can make the denominator simpler at the cost of complicating the numerator.

Your expression then becomes:

[itex]\displaystyle \frac{(x^6+x^5-x-1)(x^{3} + 5x^{2} − 9x + 4)^{3/4}}{x^5(x^{5} − x^{-5})^{3/2}}[/itex]

I don't know that this is all that much better to work with.

Added in Edit:

Let  [itex]\displaystyle w=\frac{x^6+x^5-x-1}{x^5}=x+1-x^{-4}-x^{-5}\,,[/itex]  [itex]u=x^{3} + 5x^{2} − 9x + 4\,,[/itex] and [itex]v=x^{5} − x^{-5}\,.[/itex]

Then the expression becomes: [itex]\displaystyle w\,\left(\frac{u}{v^2}\right)^{3/4}[/itex]

The derivative of this w.r.t. x is: [itex]\displaystyle \left(\frac{v^2}{u}\right)^{1/4}\cdot\left(w\cdot\left( \frac{u'\,v-v'\,u}{v^3} \right)+w'\cdot\left(\frac{u}{v^2}\right)\right)[/itex]

It would be more than a good idea to check my algebra, etc.
 
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Hm I think that might just complicate things a bit more. I'll try to juggle around with what I had earlier. Thanks, though!

Still open to any input!
 
  • #6
SammyS
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Hm I think that might just complicate things a bit more. I'll try to juggle around with what I had earlier. Thanks, though!

Still open to any input!
A plus to that last result of mine is that all of the product and quotient rules have been applied by that point. Just u, v, and w are simply polynomials --- well w does have two terms with negative exponents. Their derivatives are easy to evaluate.

I used WolframAlpha to check my expression, [itex]\displaystyle \frac{(x^6+x^5-x-1)(x^{3} + 5x^{2} − 9x + 4)^{3/4}}{x^5(x^{5} − x^{-5})^{3/2}}[/itex] against the original. Their difference is zero, so that's good.


.
 
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