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Math Help - derivative question

  1. #1
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    derivative question

    find g'(2) if g(x)=x^3h(cos(pi*x)), where h(1) = 2 h'(1)=-2

    im really stuck on this one and frankly don't even know where to start

    thanks so much
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  2. #2
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    The standard method to find the derivative of functions of type f^g where f and g are functions of x is by taking logarithm on both sides and proceeding.
    On the right side you have this very occurrence. Here, f = x while g = 3h(\cos \pi\cdot x).
    Let's take log both sides and proceed...

    \log g(x)=3h(\cos \pi\cdot x) + \log(x).
    Differentiating w.r.t. x throughout...
    1/g(x) \cdot g(x)^\prime=3h^\prime(\cos \pi\cdot x) \cdot -\pi\sin(\pi\cdot x)+1/x\quad(2)

    In your original equation if we put x=2,
    g(2)=2^{3h(cos(\pi \cdot 2))} =2^{3h(\cos 0)}=2^{3h(1)}=2^{3\times-2} which is not 0 so we can put x=2 in the eqn 2 abv.
    Which if we do, the right side of eqn 2 takes the value 1/2.

    So, we have, g^\prime(2)/g(2)=1/2.
    So, you can find g^\prime(2), now
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  3. #3
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    what happened to the x^3 though infront of the 3h(cos(pi*x))
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  4. #4
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    I thought the problem was g(x)=x^{3\cdot h(\cos \pi\cdot x)} while it is actually:
    g(x)=x^3\cdot h(\cos \pi\cdot x)\quad(*)
    Better if you had used Latex or I had interpreted you better.
    Anyway,
    you can simply differentiate that equation w.r.t. x throughout and put x=2 in the equation that follows.
    Let's do that...
    d(g(x))/dx=d(x^3)/dx\cdot h(\cos \pi\cdot x) \,+\, x^3\cdot d(h(\cos \pi\cdot x))/dx
    or, g^\prime(x)=3x^2\cdot h(\cos \pi\cdot x)\,+\,x^3 h^\prime(\cos \pi\cdot x)\cdot -\pi\cdot\sin (\pi\cdot x)
    In the equation and the equation (*) above we put x=2 and get g^\prime(2)
    OK, now?
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  5. #5
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    thank you very much, it was actually a simple problem :S just that h(1)=2 confused me but now i understand.. thank you

    oh by the way my final answer was g'(2)=8 is that correct?

    also sorry about not using latex when i have some more time ill figure it out

    thanks again
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  6. #6
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    Guess its 24, my young friend!
    g^\prime(2)=3\cdot2^2\cdot h(\cos \pi\cdot 2)\,+\,2^3 h^\prime(\cos \pi\cdot 2)\cdot -\pi\cdot\sin (\pi\cdot 2)
    or g^\prime(2)=12\cdot h(1)\,+\,8 h^\prime(1)\cdot -\pi\cdot 0=2\cdot12=24
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