Even though these problems use the chain rule, I think my real problem isn't the differentiation, but the math used to simplify the answers to what the book shows. It's very frustrating.
Here is the first problem. Differentiate:
y = (2x-5)^4 (8x^2-5)^-3
I get to here (though I may have erred):
y' = -48x(2x-5)^4(8x^2-5)^-4 + 8(8x^2-5)^-3(2x-5)^3
but I can't find my way to the book's answer:
y' = 8(2x-5)^3(8x^2-5)^-4(-4x^2 + 30x -5)
Good luck folks! I think my solution is equivalent to the correct answer (feel free to check this with a calculator) but I can't see how to convert one into the other.
I wouldn't be worried about this except that it's happened on about 5 problems in a row now...
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2 comments:
Out of your solution
Multiply the right half of your equation
(8(8x^2-5)^-3(2x-5)^3) by
(8x^2-5)/(8x^2-5)
(which is a fancy form of "1")
(It might be easier to see if you write your equation as a ratio, with the 8x^2-5 term in the denominator.)
factor
8[(2x-5)^3] * [(8(x^2)-5)^(-4)]
(Call this Z)
You have (Z)*[(8(x^2)-5)-(2x-5)(6x)]
That is (Z) *[(8(x^2)-5)-12x+30x]
That is (Z) *(-4(x^2)+30x-5)
Q.E.D.
(Or, as my old stat prof says, "Quite Easily Done")
OK, well, that helped.
I tried your multiplication by the fancy form of 1, but I couldn't figure out how that helped with factoring. (Obviously it's some kind of trick to make factoring easier, but either I don't need it or I can't understand it.)
Today at work I had gotten what you call (Z) factored out, but then I could never figure out that 48/8 is 6 and not 16 (not even close, I know) so the math just wouldn't come out right no matter what.
I make so many math errors when I do math, it's really unreal. I was terrified on my CS3 midterm, where no calculator was allowed and I had to take a bunch of numbers mod 13. (A mod B is the remainder when you divide A by B, so for instance, 17 mod 3 is 2, and 18 mod 3 is 0.)
Thank you for your help, which made me really take another stab at this.
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