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Implicit Differentiation, chain rule, and simplifying

  1. Oct 17, 2012 #1
    Okay so, I am having trouble figuring out what exactly to do in implicit differentiation and usage of the chain rule. Like, I keep getting the wrong answer somehow. See, from what I understand you have to find the derivative of both sides then use the chain rule or something and then solve for dy/dx? Also, when I take the derivative of complex functions things become very messy and sometimes I can't figure out if I need to simplify or leave in factored form. Especially in regards to quotient derivatives. I need a little algebra reminder, I can't remember... if I have something like 2y/2x can I just cancel out the twos?(And I know I can't cancel unless they are like terms but 2 and 2 are constants and are therefore like terms?) Also, what is a rational expression because I remember that it is illegal to cancel if there are plus and minus signs? Anyway, sometimes the entire simplification is insane because of all the stuff in the derivative. Any suggestions, I might need to clarify if need be, it's hard to explain like this.
     
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  3. Oct 18, 2012 #2

    arildno

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    1. First off:
    Remember that the concept "term" means those sub-expressions separated by a plus or minus sign.
    For example, the primary terms in 2*y+(x-1/2y) are "2*y" and (x-1/2y) (furthermore, the last parenthesis contains secondary, sub-terms, "x" and "1/2y")

    "Factors" are expressions that, WITHIN a given term are separated by a multiplication sign or a division sign. In your first term, "2" and "y" are your factors. The second term, (1-1/2y) consists of itself, i.e has only one factor, (1-1/2y). For the expression WITHIN the parenthesis, the first sub-term, "1" has a single factor, while 1/2y contains two factors "1" and "2y"

    2. Now, you are allowed in a fractional expression allowed to cancel COMMON FACTORS.
    Common factors are ONLY common if they appear as factors in ALL terms, both in the numerator and denominator.
    Thus, in 2x/2y, that is (2*x)/(2*y), "2" is a common factor in all terms, and you can cancel it, so that you get x/y as simplified expression.

    In: (x*y-2*y*x*x)/(3x-y*x), we see that "x" is a common factor appearing, and thus the expression can be simplified to: (y-2*y*x)/(3-y).

    3. You really should brush up your algebra skills; you'll stagnate if you remain as uncertain about this in the future.
     
  4. Oct 18, 2012 #3
    Oh okay, I see. And, I am in Intermediate Algebra so there may be a few things I don't know but I know I covered simplification and factoring and stuff. See, I am independently studying calculus so that I'll be ready when I get to it, but I probably have no hope of testing out of trig or advanced algebra but I need the background anyway. Although I have looked at all of the content within my algebra textbook including exponential functions and logs, 3D graphs, systems, inverses, quadratics, factoring review, simplification review, inequalities, sums and series, trig functions, arcsin etc, and you know all that good stuff. I feel like I know that stuff, but I will need to build upon it. I've also practiced with plane trigonometry and unit circle and stuff like that. Anyways, I am studying mathematics mostly by myself but I understand the language so it isn't too hard to apply formulas and stuff without someone spoonfeeding, until I get stuck on something.
    So, say the derivative of function f(x), is [itex]\frac{dy}{dx}[/itex] = [itex]\frac{8x^{2} -4x + y}{2x^{3}y +4x^{4} +3}[/itex]

    Then can that be simplified further? I think I can cancel the the x's out, but they are seperated like a rational expression, I'm gonna look up the definition of rational expression real quick, anyway, that is my only doubt, the simplification afterwards.

    EDIT: Oops, I reread what you said. So they have to be factors that appear in ALL terms in both numerator and denominator... Then I just answered my question! That above can't be simplified further since it has a 3 and a y but everything else is common haha
     
    Last edited: Oct 18, 2012
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