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Differentiation involving Sin(x) as a power

  1. Sep 27, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the derivative of y=(x^2)^sinx; using the chain rule.


    2. Relevant equations

    No other relevant equations.


    3. The attempt at a solution

    I attempted to apply the Chain rule: dy/dx = dy/du X du/dx
    Subbing u for x^2, which made y = u^sinx

    I ended up with du/dx = 2x;

    and dy/du = ln(u)*cos(x)*(U)^sinx

    I know this isn't right, and all i've spent the past hour doing is confusing myself. I've searched my text book, and online, and can't find anyting (that I understand) to help me.
    To be honest, i'm not even sure how i'm supposed to apply the chain rule

    Any help would be appreciated, thank you.
     
  2. jcsd
  3. Sep 27, 2012 #2

    Mute

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    The problem with the way you have attempted it is that when you computed dy/du you did not write y entirely in terms of u, you left it in terms of u and x, but u and x are not independent. This is why your calculation isn't working you. You should replace x with ##\sqrt{u}## to do it properly.

    There is a slightly easier way to do this, as well. Note that you can write

    ##x^2 = e^{\ln (x^2)} = e^{2\ln |x|}.##

    Then, recall that ##(a^b)^c = a^{bc}## for real numbers a, b and c. This will give you a form that is easier to differentiate. (You'll want to use a different chain rule substitution, though).
     
  4. Sep 28, 2012 #3

    SammyS

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    I can see a couple of ways to do this with the chain rule. One uses the multi-variable chain rule. The other require rewrite the function.

    Multi-variable chain rule:
    Let u(x)= x2 and v(x) = sin(x)

    Then [itex]y=f(x,\,y)=u^v\,.[/itex]

    Then use [itex]\displaystyle \frac{dy}{dx}=\frac{\partial y}{\partial u}\frac{du}{dx}+\frac{\partial y}{\partial v}\frac{dv}{dx}\ .[/itex]

    The single variable method:
    [itex]\displaystyle y=(x^2)^{\sin(x)}=e^{2\ln(x)\sin(x)}\ .[/itex]

    Now apply the chain rule. You will also need the product rule.​
     
  5. Sep 28, 2012 #4

    HallsofIvy

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    Simpler, I think, than writing [itex]f(x)^{g(x)}[/itex] as [itex]e^{g(x)ln(f(x))}[/itex] is to start with [itex]y(x)= f(x)^{g(x)}[/itex] and take the logarithm of both sides: ln(y)= g(x)ln(f(x)).

    Here, the function is [itex]y(x)= (x^2)^{sin(x)}= x^{2sin(x)}[/itex]. Then [itex]ln(y)= 2sin(x) ln(x)[/itex]. Then you can use the product rule to differentiate the right side using the product rule. Of course, the derivative of ln(y), on the left, is (1/y)dy/dx. After taking the derivative of both sides multiply through by [itex]y= (x^2)^{sin(x)}[/itex] to find dy/dx.
     
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