Sunday, May 21, 2006

Calculus Problem #2: Chain Rule Confusion

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.

rvman said...

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")

Tam said...

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.