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Math Help - Differentiate each - please help

  1. #1
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    Differentiate each - please help

    Identify any general rules of calculus that you use

    i) f(x) = tan(3x)In(sin(3x))

    ii) f(x) = e^-sqrt(x) divided by x^3 +1
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  2. #2
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    For the first, use the product rule:

    tan(3x)ln(sin(3x))

    First times derivative of second plus second times derivative of first.

    tan(3x)\frac{d}{dx}[ln(sin(3x))]+ln(sin(3x))\frac{d}{dx}[tan(3x)]

    tan(3x)(3cot(3x))+ln(sin(3x))(3sec^{2}(3x))

    I'll let you tidy it up a bit.
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  3. #3
    Super Member malaygoel's Avatar
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    Here is second.
    Your function is \frac{e^{-\sqrt{x}}}{x^3 + 1}
    Its differentiation is(using quotient rule)
    \frac{3x^2(e^{-\sqrt{x}})-(e^{-\sqrt{x}})(\frac{-1}{2\sqrt{x}})}{(x^3 + 1)^2}
    It may be useful to you:
    If u,v,w,... are functions of x then
    \frac{d(uvw...)}{uvw...}=\frac{du}{u} + \frac{dv}{v} + \frac{dw}{w} + ...


    Keep Smiling
    Malay
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  4. #4
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    Hello, fair_lady0072002!

    Identify any general rules of calculus that you use

    1)\;\;f(x) \,= \,\tan3x\cdot\ln(\sin3x)

    We have: . f'(x) \;= \;\tan3x\cdot[\ln(\sin3x)]' + [\tan3x]'\cdot\ln(\sin3x) . . . product rule

    . . . . . . . . f'(x) \;= \;\tan3x\cdot\left[\frac{\cos3x}{\sin3x}\cdot3\right] \,+ \,\left[3\sec^23x\right]\cdot\ln(\sin 3x) . . . chain rule

    Simplify: . f'(x)\;=\;\underbrace{\tan 3x\cdot\cot3x}\cdot3 \,+ \,3\sec^23x\cdot\ln(\sin 3x)
    . . . . . . . . . . . . . . reciprocals

    Therefore: . f'(x)\;=\;3 + 3\sec^23x\ln(\sin3x)



    2)\;\;f(x) \,= \,\frac{e^{-\sqrt{x}}}{x^3+1}

    This one is truly messy!

    We have: . f(x)\;=\;\frac{e^{-x^{\frac{1}{2}}}}{x^3 + 1}

    Then: . f'(x) \;= \;\frac{(x^3+1)\left[e^{-x^{\frac{1}{2}}}\right]' - \left(e^{-x^{\frac{1}{2}}}\right)\left[x^3+1]'}{(x^3+1)^2} . . . quotient rule

    f'(x) \;= \;\frac{(x^3+1)e^{-x^{\frac{1}{2}}}\left(-\frac{1}{2}x^{-\frac{1}{2}}\right) \,- \,\left(e^{-x^{\frac{1}{2}}}\right)(3x^2)}{(x^3+1)^2} . . . chain rule


    Now we simplify . . . (do we hafta?)

    We have: . f'(x) \;= \; \frac{-\frac{x^3+1}{2e^{x^{\frac{1}{2}}}x^{\frac{1}{2}}} - \frac{3x^2}{e^{x^{\frac{1}{2}}}}}{(x^3 + 1)^2}

    Multiply top and bottom by 2e^{x^{\frac{1}{2}}}x^{\frac{1}{2}}:\;\;f'(x) \;= \;\frac{-(x^3+1) - 3x^2\cdot2x^{\frac{1}{2}}}{2e^{x^{\frac{1}{2}}}x^{  \frac{1}{2}}(x^3 + 1)^2}

    Therefore: . f'(x) \;= \;-\frac{x^3 \,+ 1 + 6x^2\sqrt{x}}{2e^{\sqrt{x}}\sqrt{x}(x^3+1)^2}


    [Disclaimer: not responsible for egregious errors.]

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  5. #5
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    Thanx a lot. I really appreciate your assistance.
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