derivation using chain rule or not

Discussion in 'Math' started by kokkie_d, Jan 19, 2011.

  1. kokkie_d

    Thread Starter Active Member

    Jan 12, 2009
    72
    0
    Hi,

    The following function is given:
    <br />
y(x) = \frac{1}{(b-ax)}<br />
    I want to find the derivative of this one but I am stuck between two methods:

    <br />
y(x) = \frac{1}{(b-ax)}<br />
y(x) = (b-ax)^{-1}<br />
y(x) = b^{-1}-a^{-1}x^{-1}<br />
y'(x) = 0 - a^{-1}*-x^{-2}<br />
y'(x) = \frac{1}{a}*\frac{1}{x^{2}}<br />
y'(x) = \frac{1}{ax^{2}}<br />

    On the otherhand using the chain rule:
    <br />
y(x) = \frac{1}{(b-ax)}<br />
f(z) = \frac{1}{(z)}<br />
z(x) = b-ax<br />
f'(z) = \frac{1}{z^2}<br />
z'(x) = -a<br />
y'(x) = \frac{-a}{(b-ax)^2}<br />
<br />

    Could someone shed some light on this problem. It has to do with whether or not b-ax is seen as a composite function which I do not believe it is and as such the first derivation would be correct but if it is a composite function then the chain rule needs to be applied.

    I am not using the quotient rule since the top part is not a function of x.
     
  2. Papabravo

    Expert

    Feb 24, 2006
    10,135
    1,786
    Since the first derivation has errors in algebra it is unlikely to be correct. The second derivation appears to be correct. I am curious as to why you think (b - ax) is not a composite function.
     
  3. kokkie_d

    Thread Starter Active Member

    Jan 12, 2009
    72
    0
    Hi,

    The first derivation I am just saying that I can break up a fraction thus not composite?
    What am I doing wrong in algebra in derivation 1?
     
  4. someonesdad

    Senior Member

    Jul 7, 2009
    1,585
    141
    Put some numbers in for the symbols and see where equality breaks down. You should be able to discover the problems for yourself. Remember, the algebraic relations should be true for all numbers that don't make the denominator vanish.
     
    kokkie_d likes this.
  5. Papabravo

    Expert

    Feb 24, 2006
    10,135
    1,786
    In the third step you apply the -1 exponent to both b and ax and this is not correct.

    (b-ax)^-1 ≠ b^-1 - a^-1x^-1

    Let b=2, a=3, & x=4

    The expression on the right evaluates to -(1/10) while the expression the right evaluates to 5/12. When it comes to counter examples all I need is one!

    I'm not sure what you mean by breaking up a fraction, but whatever you were thinking the result was incorrect.

    You could also have used the division rule, whereby
    Code ( (Unknown Language)):
    1.  
    2. d(u/v) = (vdu - udv) / (v^2)
    3.   =((b-ax)(0) - (1)(-a)) / (b-ax)^2
    4.   = a / (b-ax)^2
    5.  
    This is different than your second result. You missed a minus sign due to the negative exponent that you started with.

    Code ( (Unknown Language)):
    1.  
    2. d(z^-1) = -z^-2
    3. d(-z^-2) = 2z^-3
    4. and so on
    5.  
     
    Last edited: Jan 19, 2011
    kokkie_d likes this.
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