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Math Help - logarithmic differentiation

  1. #1
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    logarithmic differentiation

    i'm so sorry for keep asking questions but im really confused for this one.
    the problem is y=(e^(x-1)*sin^2*x)/(((x^2)+5)^2x)

    the trig function (sin^2 + 5), i don't know how to get a derivative out of that.

    help me plz!
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  2. #2
    tah
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    Quote Originally Posted by katieeej View Post
    i'm so sorry for keep asking questions but im really confused for this one.
    the problem is y=(e^(x-1)*sin^2*x)/(((x^2)+5)^2x)

    the trig function (sin^2 + 5), i don't know how to get a derivative out of that.

    help me plz!
    Hi, may be there is a typo on the definition of the trigonometric function, is it sin^(x) + 5 or sin^2(x+5) or y is the variable or, ...May be I am missing the question
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  3. #3
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    the equation is

    e^(x-1)*sin^2 (x) / (x^2 + 5)^2x

    its really complicated i'm sorry!
    it's sin square of (x) and (x^2 +5) is x square and plus 5, separately.

    thanks!
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  4. #4
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    We have:
    y=\frac{e^{x-1}\sin^2{(x)}}{(x^2+5)^{2x}}={e^{x-1}\sin^2{(x)}}{(x^2+5)^{-2x}}

    To make things cleaner, we use the product rule and let:
    f(x)=e^{x-1}\sin^2(x)
    and
    g(x)=(x^2+5)^{-2x}=e^{\ln(x^2+5)^{-2x}}=e^{-2x\ln(x^2+5)}

    We now have:
    \frac{dy}{dx}=f'(x)g(x)+g'(x)f(x)


    Using the product:
    f'(x)=e^{x-1}\sin^2(x)+2\sin(x)\cos(x)e^{x-1}=e^{x-1}\sin(x)(2\cos(x)+\sin(x))

    Using the chain rule:
    g'(x)=e^{-2x\ln(x^2+5)}\frac{d}{dx}-2x\ln(x^2+5)
    =-2(x^2+5)^{-2x}\frac{d}{dx}x\ln(x^2+5)
    =-2(x^2+5)^{-2x}(x\frac{d}{dx}\ln(x^2+5)+\ln(x^2+5))
    =-2(x^2+5)^{-2x}(\frac{x}{x^2+5}(2x)+\ln(x^2+5)) here I use the property: \frac{d}{du}\ln(u)=\frac{1}{u}
    =-2(x^2+5)^{-2x}(\frac{2x^2}{x^2+5}+\ln(x^2+5))

    Now I'll leave it for you to plug in and simplify.
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  5. #5
    tah
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    Quote Originally Posted by katieeej View Post
    the equation is

    e^(x-1)*sin^2 (x) / (x^2 + 5)^2x

    its really complicated i'm sorry!
    it's sin square of (x) and (x^2 +5) is x square and plus 5, separately.

    thanks!
    the derivative looks a bit ugly but it follows from the formulas
    (uv)' = u'v + uv'\ and\ (\frac{u}{v})' = \frac{u'v-uv'}{v^2}
    which I am sure you have already used.

    Try it out and compare your result to

     \frac{e^{x-1}sin(x) (x^3 sin(x) - 5 x^2sin(x) + 5 x sin(x) - 5 sin(x) + 2 x^3cos(x) + 10 x cos(x))}{x ^2(x ^2+ 5)^3 }

    which was done by a symbolic manipulation software (Yes I didn't do it )
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  6. #6
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    Differentiating

    Hello katieeej
    Quote Originally Posted by katieeej View Post
    the equation is

    e^(x-1)*sin^2 (x) / (x^2 + 5)^2x

    its really complicated i'm sorry!
    it's sin square of (x) and (x^2 +5) is x square and plus 5, separately.

    thanks!
    I'm sure you'll find it easier to begin by taking logs:

    ln(y) = (x-1) + 2ln(\sin x) - 2x \cdot ln(x^2+5)

    Then differentiate:

    \frac{1}{y}\frac{dy}{dx} = ... etc.

    and finally multiply both sides by y (= ...original function of x).

    Grandad
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