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Differentiate e^(-1/2 x) sqrt(1 + 2x) - troubles!

  1. Aug 4, 2010 #1
    the curve y = e^ −1/2x √(1 + 2x) please differentiate this equation i am having some problem.. i want to check my answer

    My answer :

    e^-1/2x { (1+2x )^-1/2 + (1+2x)^1/2 (-1/2) }

    In fact i need to find the x-coordinates and according the booklet the answer of the x-coordinate is 0.5 and me i cannot have the answer ( need to equal the dy/dx = 0 ) to find x-coordinate
     
  2. jcsd
  3. Aug 4, 2010 #2

    ehild

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    Re: differenciation...

    Do you need to find the value of x where dy/dx=0? You derived the equation already, put it equal to 0 and solve. As the exponent can not be zero, the factor { (1+2x )^-1/2 + (1+2x)^1/2 (-1/2) }=0.

    ehild
     
  4. Aug 4, 2010 #3

    Mentallic

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    Re: differenciation...

    differentiation of two products means you must use the product rule.

    if y=uv, where u and v are functions of x, such as in your case you have [itex]u=e^{-x/2}[/itex] and [itex]v=(1+2x)^{1/2}[/itex]

    then [tex]\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}[/tex]

    Ok so what is [tex]\frac{dv}{dx}[/tex] and what is [tex]\frac{du}{dx}[/tex] ?

    Solve these separately first and we'll show you if you made any mistakes anywhere in your differentiation.
     
  5. Aug 4, 2010 #4
    Re: differenciation...

    Okk so for U = e^-1/2x...........du/dx = -1/2 e^-1/2x

    and for V = (1+2x)^1/2..........dv/dx = (1+2x)^-1/2

    and i replace in y=uv
     
  6. Aug 4, 2010 #5

    Mentallic

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    Re: differenciation...

    Nearly. For [tex]v=(1+2x)^{1/2}[/tex] you need to remember to take the derivative of the inside of the brackets by the chain rule. For [tex]y=[f(x)]^n[/tex], [tex]\frac{dy}{dx}=n[f(x)]^{n-1}.f'(x)[/tex]

    And then you replace in [tex]\frac{dy}{dx}=uv'+vu'[/tex] where v' is the derivative of v with respect to x, or [tex]\frac{dv}{dx}[/tex] and similarly for u'.
     
  7. Aug 4, 2010 #6
    Re: differenciation...


    no i have already evalute,, 1/2 (1+2x) ^-1/2 x 2 = (1 +2x) ^-1/2
     
  8. Aug 4, 2010 #7
    Re: differenciation...


    hmmm no let the exponential and x-coordinate part,, help me to differentiate, infact i have differentiate, but i am not obtaining the answer
     
  9. Aug 4, 2010 #8
    Re: differenciation...

    yeahh infact the x-coordinate of a maximum point
     
  10. Aug 4, 2010 #9

    ehild

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    Re: differenciation...

    You have differentiate the y(x) function already. Now you have to solve the equation

    (1+2x )^-1/2 + (1+2x)^1/2 (-1/2)=0

    I rewrite it in tex so as you see more clearly:

    [tex]\frac{1}{\sqrt{1+2x}}-\frac{1}{2}\sqrt{1+2x}=0[/tex]

    hint: Multiply the equation with (1+2x) ^1/2.

    ehild
     
  11. Aug 4, 2010 #10
    Re: differenciation...

    [tex]\frac{1}{\sqrt{1+2x}}-\frac{1}{2}\sqrt{1+2x}=0[/tex]

    Till here i have reached, so

    1 / (1+2x)^1/2 = 1/2 (1+2x)^1/2

    If i use your hint i get: 0 = 1/2 (1+2X) ^1/2 x (1+2X) ^1/2

    Aww now how should i break the brackets ?
     
  12. Aug 4, 2010 #11
    Re: differenciation...

    ## Oppps i think that it is not zero but one
     
  13. Aug 4, 2010 #12

    ehild

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    Re: differenciation...

    Your last equation is totally wrong. I said multiply the whole equation by
    (1+2X) ^1/2. Do it with your first equation, it is correct.

    ehild
     
  14. Aug 4, 2010 #13
    Re: differenciation...


    so the equation is like this after the multiplication:

    1= 1/2 (1+2X)^1/2 X (1+2X) ^1/2

    1 = 1/2 (1 + 2x)

    is it correct?
     
  15. Aug 4, 2010 #14
    Re: differenciation...

    by the way the answer that is the x-coordinate is 0.5 according to the booklet
     
  16. Aug 4, 2010 #15

    HallsofIvy

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    Re: differenciation...

    Yes, that is correct. And that is a simple linear equation. Solve it for x.
     
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