Problem with Differentiation Using Quotient Rule

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SUMMARY

The discussion focuses on finding the second derivative of the function h(x) = [(x^2)-1] / [2x-(x^2)] using the Quotient Rule. The first derivative is correctly calculated as h'(x) = [2(x^2)-2x+2] / [2x-(x^2)]^2. However, the second derivative h''(x) = [2x(4(x^5)-5(x^4)+(x^3)+20(x^2)-16x+4)] / [2x-(x^2)]^4 is deemed incorrect and overly complex. Simplification is necessary, particularly by factoring out (2x-x²) to streamline the result.

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I am attempting to find the second derivative of a function:

h(x) = [(x^2)-1] / [2x-(x^2)]

I proceeded by using the Quotient Rule, and I found the following as the first derivative. (It is correct.)

h`(x) = [2(x^2)-2x+2] / [2x-(x^2)]^2

Next, I tried using the Quotient Rule again, and I found a very long result. I simplified the result and obtained the following:

h``(x) = [2x(4(x^5)-5(x^4)+(x^3)+20(x^2)-16x+4)] / [2x-(x^2)]^4

Is this correct? And, if so, how can I proceed from here? (The answer in the book seems to have been simplified a lot more than this one.

Thanks in advance!
Nico
 
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Not correct, I'm afraid. You'll need to check your working.

One way of checking if your answers are correct is to graph them, using a graphing calculator or software.

h'() is the slope of h(), and h''() is the slope of h'().
If you plot h'(), you will find it has a minimum at about x = 0.8. That means its slope, the h''() curve, is zero at about x = 0.8. So if you plot h''(), and it cuts the x-axis at about x = 0.8, then that is reasonable confirmation that you may have the right answer.
Also, when doing the 2nd derivative, you should get a common factor of (2x-x²) that cancels out top and bottom. This will simplify things a bit.
 

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