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Not certain about derivative.

  1. Mar 13, 2007 #1
    1. The problem statement, all variables and given/known data

    If [tex]y=\frac{1-2u}{1+u}[/tex] and [tex]u=\sqrt{x^2-7}[/tex], find [tex]\frac{dy}{dx}[/tex] at [tex]x=4[/tex]

    2. Relevant equations

    quotient and chain rule

    3. The attempt at a solution

    so if [tex]f=y[/tex] then [tex]y^{'}=f^{'}[/tex] then [tex]f^{'} = \left( \frac{f}{g}\right)^{'} = \frac{f^{'}g - fg^{'}}{g^2}[/tex]

    and if [tex]f=1-2u[/tex] then [tex]f^{'} = -\left( x^2 -7\right)^\frac{-1}{2} 2x[/tex]

    and if [tex]g=1+u[/tex] then [tex]g^{'} = x\left( x^2 -7\right)^\frac{-1}{2}[/tex]

    and [tex]g^2 = x^2 + 2\sqrt{x^2 -7}-6[/tex]

    and now with chain rule:


    i think?
    u in terms of x like this:

    Last edited: Mar 14, 2007
  2. jcsd
  3. Mar 13, 2007 #2
    Try the chain rule:

    dy/dx =dy/du(du/dx)

    should be easy as that
  4. Mar 13, 2007 #3


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    Homework Helper

    your f ', g ' and g^2 look correct
  5. Mar 13, 2007 #4
    You could also simply substitute u(x) back into the equation, though I would personally go with chain rule.
  6. Mar 13, 2007 #5
    okay, hopefully i can simplify it somehow...
  7. Mar 14, 2007 #6


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    Science Advisor

    Did you understand what was said? You appear to be completely ignoring it. YOU put the "chain rule" as one of the relevant equations. The chain rule is specifically to do problems like this without having to substitute one function into another. Use the chain rule, not that complicated formula for y(x) you have!
  8. Mar 14, 2007 #7
    Okay, i will try it again with the chain rule, i dont understand how the chain rule can be used since i cant see a function thats "in" another function? where is it? i actually made a mistake when i wrote chain rule, but since you say its easier i want to do it that way, i will search for a suitable example.

    later that day....

    ah, now i see, you don't substitute for u as i did, i also see the solution will be much cleaner, thanks!
    Last edited: Mar 14, 2007
  9. Mar 14, 2007 #8
    All the mathematicians will probably hate this, but just think of d/dx as something that you want to "cancel" out. So the chain rule says that if you have h(g(f(x))), which you can repeat indefinitely, that you can take the derivative of dh/dx by incrementally "canceling" the derivatives, like so dh/dx = dh/dg*dg/df*df/dx.

    So as an example, the say you have [tex] y = sin((x+1)^2) [/tex], which has a form y = h(g(f(x))), where h is the trig function sin(stuff), g is the function inside of the trig function raised to some power (stuff)^2, and f is the function x+1. The solution would be
    [tex]\frac{d(sinfunction)}{dg} * \frac{d(powerfunction)}{df} * \frac{d(finalfunction)}{dx} [/tex]
    which evaluates to cos((x+1)^2)*2(x^50+1)*50x^49.

    You have a function in terms of u, which is also a function of x. So with the chain rule the equation
    [tex] y(u) = \frac{1-2u}{1+u}[/tex] where [tex]u(x) = \sqrt{x^2-7}[/tex]

    has the form of [tex]y(h) = \frac{1-2h(g(f(x)))}{1+h(g(f(x)))}[/tex]
    *remember there is a square root function in u's expression for x, which is a function in itself.

    so dy/dx = dy(h)/dg(f)*dg(f)/df(x)*df(x)/dx.

    Make sense?
    Last edited: Mar 14, 2007
  10. Mar 14, 2007 #9
    makes perfect sense.
  11. Mar 14, 2007 #10
    Catch my edit though, I forgot there was a square root function in the function for u(x).
  12. Mar 14, 2007 #11
    yeah i see the change there too.
  13. Mar 14, 2007 #12
    Yes, you seem to have caught the sqrt, so your new answer is right. I would put u in terms of x now though.
  14. Mar 14, 2007 #13
    is that what you mean when you say u in terms of x?
  15. Mar 14, 2007 #14
    Yeah, it's nice to have everything in terms of the variable you a taking the derivative with respect to. :)
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