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Math Help - Differentiation

  1. #1
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    Differentiation

    Hi,

    The question is: Differentiate the following function, identifying any general rules of calculus that you use and simplify your answer as far as possible.

    f(x)=\frac{e^{tanx}}{(x^3+1)}

    Rules: quotient rule, product rule, sum/multiple rules

    forulma: f'(x)=\frac{g(x)e'(x)-e(x)g'(x)}{(g(x))^2}

    e(x)=e^{tanx}, u=tanx
    e'(u)=e^u, \frac{du}{dx}=sec^2x
    \frac{de}{du}=e^u
    \frac{de}{dx}=e^u x sec^2x=sec^2 x e^{tanx}

    g(x)=(x^3+1)^2, u=x^3+1
    g'(u)=u^2, \frac{du}{dx}=3x^2
    \frac{dg}{du}=2u
    \frac{dg}{dx}=2(x^3+1)3x^2

    f'(x)=\frac{(x^3+1)^2 x (sec^2x e^{tanx})-(e^{tanx})x(2(x^3+1)3x^2)}{((x^3+1)^2)^2}
    f'(x)=\frac{(x^3+1)^2 x sec^2x e^{tanx}-e^{tanx}x2(x^3+1)3x^2}{(x^3+1)^4}

    I hope this is right so far, but am not sure how to simplify any further (unless I've done something wrong?)
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by cozza View Post
    Hi,

    The question is: Differentiate the following function, identifying any general rules of calculus that you use and simplify your answer as far as possible.

    f(x)=\frac{e^{tanx}}{(x^3+1)}

    Rules: quotient rule, product rule, sum/multiple rules

    forulma: f'(x)=\frac{g(x)e'(x)-e(x)g'(x)}{(g(x))^2}

    e(x)=e^{tanx}, u=tanx
    e'(u)=e^u, \frac{du}{dx}=sec^2x
    \frac{de}{du}=e^u
    \frac{de}{dx}=e^u x sec^2x=sec^2 x e^{tanx}

    g(x)=(x^3+1)^2, u=x^3+1
    g'(u)=u^2, \frac{du}{dx}=3x^2
    \frac{dg}{du}=2u
    \frac{dg}{dx}=2(x^3+1)3x^2

    f'(x)=\frac{(x^3+1)^2 x (sec^2x e^{tanx})-(e^{tanx})x(2(x^3+1)3x^2)}{((x^3+1)^2)^2}
    f'(x)=\frac{(x^3+1)^2 x sec^2x e^{tanx}-e^{tanx}x2(x^3+1)3x^2}{(x^3+1)^4}

    I hope this is right so far, but am not sure how to simplify any further (unless I've done something wrong?)
    I disagree with something here...if f(x)=\dfrac{e^{\tan x}}{x^3+1}, then by your method, e(x)=e^{\tan x} and g(x)=x^3+1 (not (x^3+1)^2). As a result, you're quotient rule should give you f^{\prime}(x)=\dfrac{(x^3+1)e^{\tan x}\sec^2x-3x^2e^{\tan x}}{(x^3+1)^2}=\dfrac{e^{\tan x}\left[(x^3+1)\sec^2 x-3x^2\right]}{(x^3+1)^2}, which really can't be simplified further....
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  3. #3
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    Quote Originally Posted by Chris L T521 View Post
    I disagree with something here...if f(x)=\dfrac{e^{\tan x}}{x^3+1}, then by your method, e(x)=e^{\tan x} and g(x)=x^3+1 (not (x^3+1)^2). As a result, you're quotient rule should give you f^{\prime}(x)=\dfrac{(x^3+1)e^{\tan x}\sec^2x-3x^2e^{\tan x}}{(x^3+1)^2}=\dfrac{e^{\tan x}\left[(x^3+1)\sec^2 x-3x^2\right]}{(x^3+1)^2}, which really can't be simplified further....

    Sorry, the original question shoul have said (x^3+1)^2, not (x^3+1)
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  4. #4
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    Quote Originally Posted by cozza View Post
    Hi,

    The question is: Differentiate the following function, identifying any general rules of calculus that you use and simplify your answer as far as possible.

    f(x)=\frac{e^{tanx}}{(x^3+1)}

    Rules: quotient rule, product rule, sum/multiple rules

    forulma: f'(x)=\frac{g(x)e'(x)-e(x)g'(x)}{(g(x))^2}

    e(x)=e^{tanx}, u=tanx
    e'(u)=e^u, \frac{du}{dx}=sec^2x
    \frac{de}{du}=e^u
    \frac{de}{dx}=e^u x sec^2x=sec^2 x e^{tanx}


    where did the factor "x" come from?


    g(x)=(x^3+1)^2, u=x^3+1
    g'(u)=u^2, \frac{du}{dx}=3x^2
    \frac{dg}{du}=2u
    \frac{dg}{dx}=2(x^3+1)3x^2

    f'(x)=\frac{(x^3+1)^2 x (sec^2x e^{tanx})-(e^{tanx})x(2(x^3+1)3x^2)}{((x^3+1)^2)^2}
    f'(x)=\frac{(x^3+1)^2 x sec^2x e^{tanx}-e^{tanx}x2(x^3+1)3x^2}{(x^3+1)^4}

    I hope this is right so far, but am not sure how to simplify any further (unless I've done something wrong?)

    f(x)=\displaystyle\huge\frac{e^{tanx}}{(x^3+1)^2}=  e^{tanx}\frac{1}{(x^3+1)^2}=e^{tanx}\left(x^3+1\ri  ght)^{-2}

    You can use the product rule...

    f'(x)=\displaystyle\huge\frac{1}{(x^3+1)^2}\left(e  ^{tanx}sec^2x\right)+e^{tanx}\left(-2\frac{1}{(x^3+1)^3}3x^2\right)

    then clean that up..
    Last edited by Archie Meade; August 2nd 2010 at 08:55 AM. Reason: typo
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