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Tricky Looking!

  1. May 11, 2005 #1
    y = (4-x) ^ 5x

    Just wondering what i should do with this
    Thanx
     
  2. jcsd
  3. May 11, 2005 #2
    I guess that depends what the question is. Are you asking for the differential equation to which this is the solution?

    Take logs and differentiate for first order ODE?
     
  4. May 12, 2005 #3
    yes, i have to differentiate it

    sorry i didnt specifiy what had to be done, i presumed you knew it was differentiation.
     
  5. May 12, 2005 #4
    If it's just finding the derivative then this is probably more a question for the calculus thread.

    Anyhoo, i'll do the first part. Take logs to get.

    [tex] ln(y)= 5x ln(4-x) [/tex]

    Now differentiate (implicitly on the LHS) and rearrange to get your answer for dy/dx. Have a go and let me know if/where you get stuck.
     
  6. May 12, 2005 #5
    ok, i understand your instructions, its just a matter of doing the right things now
    This is what i got as an answer, unfortunately we arent supplied with answers in this exercise.

    following on from your step, this is what i did

    1/y * dy/dx = 5 * (1/4-x)
    dy/dx = (5/4-x)y
    dy/dx = (5/4-x)(4-x)^5x
     
  7. May 12, 2005 #6

    dextercioby

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    Pay attention with the differentiaition of the RHS.It's a product.I'm sure one of the 2 terms will contain the natural logarithm.

    Daniel.
     
  8. May 12, 2005 #7
    Think with my poor Tex you missed the x after the 5 on the RHS. Take that into account and you're there.
     
  9. May 12, 2005 #8

    arildno

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    Here's another way:
    Let [tex]y(x)=f(u(x),v(x)), f(u,v)=u^{v}, u(x)=4-x, v(x)=5x[/tex]
    Then, we have:
    [tex]\frac{dy}{dx}=\frac{\partial{f}}{\partial{u}}\frac{du}{dx}+\frac{\partial{f}}{\partial{v}}\frac{dv}{dx}[/tex]
     
  10. May 13, 2005 #9
    ahh i see. i didnt spot the product rule on the right, i guess practice makes perfect.
    Thanx everyone for the input.
     
  11. May 15, 2005 #10
    when differentiating functions of a variable say x raised to another function of xthen lets assume they are T^u,where t and u are fucntions of x,dy/dx =
    T^u[du/dx*logt+u(dt/dx)/t]
     
  12. May 15, 2005 #11
    when differentiating functions of a variable say x raised to another function of xthen lets assume they are T^u,where t and u are fucntions of x,dy/dx =
    T^u[du/dx*logt+u(dt/dx)/t].now let 4-x=t,and u=5x.
    take normal procedures and see if it works
     
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