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Thread: Derivatives#2

  1. #1
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    Derivatives#2

    y=(1+x^2)tan^-1(2x)
    Yet again I am stuck on this one. I am studying for a test next week. I am not sure where to even start on this. Can someone work this one complete?
    Thanks,
    Caleb
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  2. #2
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    Here's a hint: by the Product Rule,

    $\displaystyle \frac{dy}{dx}=2x\cdot\tan^{-1}(2x)+(1+x^2)\cdot\frac{d}{dx}\tan^{-1}(2x)\cdot 2,$

    and the derivative of $\displaystyle \tan^{-1}x$ is $\displaystyle \frac{1}{1+x^2}$.
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  3. #3
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    I take it the problem is:

    $\displaystyle y= [1+x^{2}] * [ \arctan 2x] $?

    Use the product rule.

    Hint: the derivative of $\displaystyle arctan(u)$ is $\displaystyle \frac {u'}{1+u^{2}}$
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  4. #4
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    Ahhh,
    Yes it is arctan.
    Thanks Guys for all of your help. I am just not understanding derivatives at all, and you guys have been very helpful in me trying to learn them!
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  5. #5
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    Like anything, they get easier with practice.
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  6. #6
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    I scanned this particular problem, since I didnt know how to type it in. Any help on it?
    Attached Thumbnails Attached Thumbnails Derivatives#2-cmrgfix-005.jpg  
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  7. #7
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    Since $\displaystyle y= ln (x^{n})$ = $\displaystyle n \ln (x) $, then:

    $\displaystyle \ln [\frac {3x+2}{3x-2}]^{ \frac {1}{3}} $ = $\displaystyle y= \frac {1}{3} \ln \frac {3x+2}{3x-2}$

    Since you are dividing, break it up into two natural logarithms.

    $\displaystyle y= \ln \frac {a}{b}$ = $\displaystyle \ln a - \ln b$

    Does that help? Just use the rules of logarithms to simply so that you can find the derivative.

    Hint:
    if $\displaystyle y= \ln u$, then $\displaystyle y'= (\frac {1}{u}) \frac {du}{dx} $
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