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Easy chain rule for you guys

  1. Mar 24, 2008 #1
    [tex] y=2x^{sinx}[/tex]

    i know i should use the product rule within a chain rule. but how can i use chain rule with sinx

    is the anwser

    [tex] y=-2x^{cosx}[/tex]

    can anyone give me pointer to this easy problem and tell if am forgetting something.
     
  2. jcsd
  3. Mar 24, 2008 #2
    or maybe the answer is

    [tex] y=-2cosx [/tex]
     
  4. Mar 24, 2008 #3
    That definitely is not the answer. You need to realize what you're doing when you apply the chain rule. First of all, you are attemping to find [itex]\frac{dy}{dx}[/itex], which you have not indicated. Show your work and maybe we can help, but you first have to transform the equation into something you can work with (i.e. get rid of the exponent). How would you accomplish this?

    Edit: Is your equation [itex]y=2xsin(x)[/itex] or [itex]y=2x^{sinx}[/itex]?
     
  5. Mar 24, 2008 #4
    [itex]\frac{dy}{dx}[/itex] [tex] {y= 2x^{sinx}} [/tex]

    product rule
    [tex] f= 2 f'= 0
    g= x g'=? [/tex]
    chain rule
    [tex] f= x f'= 1
    g=? g'=? [/tex]
    is the anser
    [tex]
    y'=-2x^{cosx} [/tex]
     
    Last edited: Mar 24, 2008
  6. Mar 24, 2008 #5
    You're making the assumption that:

    [tex]\frac{d}{dx}\;2x^{sin(x)}=-2x^{cos(x)}\rightarrow (f \circ g)' = f'\circ g\cdot g'=-2x^{cos(x)}[/tex]

    It doesn't.

    You're using the chain rule but incorrectly, use the product rule on the results.
     
    Last edited: Mar 24, 2008
  7. Mar 24, 2008 #6

    cristo

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    Staff Emeritus
    Science Advisor

    To evaluate this derivative I would take logarithms of both sides to get ln(y)=sinx.ln(2x) and then differentiate implicitly.
     
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