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Thread: 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.

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

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

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

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

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

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

    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.

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

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

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

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

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

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

    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 $\displaystyle f(x)=\dfrac{e^{\tan x}}{x^3+1}$, then by your method, $\displaystyle e(x)=e^{\tan x}$ and $\displaystyle g(x)=x^3+1$ (not $\displaystyle (x^3+1)^2$). As a result, you're quotient rule should give you $\displaystyle f^{\prime}(x)=\dfrac{(x^3+1)e^{\tan x}\sec^2x-3x^2e^{\tan x}}{(x^3+1)^2}=\dfrac{e^{\tan x}\left[(x^3+1)\sec^2 x-3x^2\right]}{(x^3+1)^2}$, 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 $\displaystyle f(x)=\dfrac{e^{\tan x}}{x^3+1}$, then by your method, $\displaystyle e(x)=e^{\tan x}$ and $\displaystyle g(x)=x^3+1$ (not $\displaystyle (x^3+1)^2$). As a result, you're quotient rule should give you $\displaystyle f^{\prime}(x)=\dfrac{(x^3+1)e^{\tan x}\sec^2x-3x^2e^{\tan x}}{(x^3+1)^2}=\dfrac{e^{\tan x}\left[(x^3+1)\sec^2 x-3x^2\right]}{(x^3+1)^2}$, which really can't be simplified further....

    Sorry, the original question shoul have said $\displaystyle (x^3+1)^2$, not $\displaystyle (x^3+1)$
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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.

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

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

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

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


    where did the factor "x" come from?


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

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

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

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

    You can use the product rule...

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

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