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Derivative of (sin(x))^x

  1. Apr 5, 2004 #1
    What are the steps used to obtain the derivative of (sin(x))^x? I know it's (sin(x))^x [xcot(x) + ln(sinx)] however I don't know how to get there.
     
  2. jcsd
  3. Apr 6, 2004 #2
    Ahhhh a "tough" question. Most texts will show the proof for [itex]\frac{d}{du}a^u[/itex] and [itex]\frac{d}{du}u^a[/itex] where a is a constant and u is a function. What you have is a combination of the two where the base is a function raised to the power of a function.

    Lets look at a simple proof that should help you with the above.

    [tex]y=U^V[/tex] where u and v are variables.

    [tex]\ln{y}=\ln{U^V}[/tex] we'll do a little log magic.

    [tex]\ln{y}=V\ln{U}[/tex] remember how we deal with powers within log function.

    [tex]\frac{d}{dx}(\ln{y}=V\ln{U})[/tex] we'll do some implicit differentiation.

    [tex]\frac{dy}{dx}\frac{1}{y} = V\frac{1}{U}\frac{dU}{dx} + \frac{dV}{dx}\ln {U}[/tex] remember your chain rule.

    [tex]\frac{dy}{dx} =y (V \frac{1}{U} \frac{dU}{dx} + \frac{dV}{dx} \ln {U})[/tex] move our y over remember [itex]y=U^V[/itex].

    [tex]y^\prime =U^V (V U^{-1} U^\prime + V^\prime \ln {U})[/tex] change our notation to make things look cleaner.

    thus we have
    [tex]y^\prime = V U^{V-1} U^\prime + U^VV^\prime \ln {U}[/tex]

    The above should give you the answer you are looking for. One recomendation if you don't do this already: for rules that require a mixture of prime and non prime functions I'd make a table like this:

    given
    [tex]y=x\sin{x}[/tex]
    [tex]\frac{d}{dx}(y=x\sin{x})[/tex] you need the chain rule here.

    let [tex] U=x [/tex] and [tex] V=\sin{x} [/tex]
    thus:[tex] U^\prime=1 [/tex] and [tex] V^\prime=\cos{x} [/tex]

    now just plug these into the chain rule [itex] y^{\prime}=UV^{\prime}+U^{\prime}V[/itex]

    Writing your information in a clean tabular format will mean you are less likely to make a mistake when plugging in the primes and non prime functions into the rule. Just a thought.

    Good Luck
     
  4. Apr 6, 2004 #3
    Of course! Differentiation by logarithms. Thanks.
     
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