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Math Help - Differential calculus

  1. #1
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    Differential calculus

    i have two questions, for both i need to find F'(x) of the function.
    i think both are using the product rule but not 100% certain. All help much appreciated.

    1 f(x)= ln(x^3-1/3x^2)

    2 f(x)=e^cosx *sin(ln x)
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  2. #2
    Super Member wingless's Avatar
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    1. Use chain rule and quotient rule.
    2. Use chain rule and product rule.
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  3. #3
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    Quote Originally Posted by ollieman View Post
    i have two questions, for both i need to find F'(x) of the function.
    i think both are using the product rule but not 100% certain. All help much appreciated.

    1 f(x)= ln(x^3-1/3x^2)

    2 f(x)=e^cosx *sin(ln x)

    1.
    f(x)= \ln \left( \frac{x^3-1}{3x^2}\right)

    Using the log rules: \ln \left(\frac{a}{b}\right) = \ln (a) - \ln(b)

    f(x)= \ln \left( x^3-1\right) - \ln (3x^2)

    \ln is given if in a fraction, the denominator differentiated gives the numerator. \ln x = \int \frac{1}{x}

    f'(x) = \frac{3x^2}{x^3-1} - \frac{6x}{3x^2}


    2.
    f(x)=e^{\cos x} *\sin(\ln x)

    u = e^{\cos x}, \ u' = -\sin x . e^{\cos x}, \ v= \sin(\ln x), \ v' = \frac{\cos(\ln (x))}{x}

    \frac{\mathrm{d}(uv)}{\mathrm{d}x} = uv' + u'v

    f'(x) = (e^{\cos x})\left(\frac{\cos(\ln (x))}{x}\right) + (-\sin x . e^{\cos x})(\sin(\ln x))
    Last edited by Simplicity; June 29th 2008 at 03:22 AM. Reason: Latex Correction
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  4. #4
    Senior Member nikhil's Avatar
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    Check this out

    Hi!ollieman
    Chain rule should be used for sureand
    I will solve one of the question
    as an example.(you may do other one by urself,ask if you have any doubt in solving)
    y=f(x)=[e^cosx]sin(ln(x))
    differentiating we get
    y'=[d/dx(e^cosx)]sin(ln(x))+[e^cosx]d/dx(sin(lnx))
    y'=[d/dx(e^cosx)]sin(ln(x))+[e^cosx]cos(lnx)d/dx(lnx)
    y'=[d/dx(e^cosx)]sin(ln(x))+[e^cosx]cos(lnx)(1/x)
    we may solve [d/dx(e^cosx)] saperately as it is a power function
    let A=e^cosx
    lnA=cosx
    1/A(dA/dx)=-sinx
    dA/dx=-sinx*A=-e^cos(x)[sin(x)]
    so finally
    y'= -[e^cos(x)]sin(x)sin(lnx)+[e^cosx]cos(lnx)(1/x)
    Last edited by nikhil; June 29th 2008 at 03:22 AM. Reason: Rearrangement of words
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