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Thread: Finding Deriv when # is raised ^x

  1. #1
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    Finding Deriv when # is raised ^x

    This is a pretty simple one I guess but I'm not understanding whats happening here two part question:

    First simply finding d/dx(4^x)?

    The ti-89 says 2*ln(2)*4^x and I'm not getting the mechanics of it.

    This is all part of a larger problem which is:

    Find f'(x): (4^x*x^(1/3))/tanx

    Do I use the product rule seperately to get the deriv of the numerator and then do the quotient rule for the new numerator/tan(x)?
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  2. #2
    Super Member redsoxfan325's Avatar
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    Let $\displaystyle 4^x = e^{ln(4^x)} = e^{xln(4)}$. Thus the derivative, using the chain rule equals: $\displaystyle e^{xln(4)}*ln(4)$. From the first equality, we know this equals $\displaystyle 4^x*ln(4)$. The TI-89 says $\displaystyle 4^x*2ln(2)$ because $\displaystyle ln(4)=ln(2^2)=2ln(2)$.

    In general, $\displaystyle \frac{d}{\,dx}c^x = ln(c)*c^x$, where $\displaystyle c$ is a constant.
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  3. #3
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    Quote Originally Posted by mattc View Post
    This is a pretty simple one I guess but I'm not understanding whats happening here two part question:

    First simply finding d/dx(4^x)?

    The ti-89 says 2*ln(2)*4^x and I'm not getting the mechanics of it.

    This is all part of a larger problem which is:

    Find f'(x): (4^x*x^(1/3))/tanx

    Do I use the product rule seperately to get the deriv of the numerator and then do the quotient rule for the new numerator/tan(x)?
    Let y = 4^x so that $\displaystyle ln(y) = x\ln{4} = 2x\ln{(2)}$

    Differentiating:
    $\displaystyle
    (\frac{1}{y})(\frac{dy}{dx}) = 2\ln{(2)}$

    multiply by y:

    $\displaystyle \frac{dy}{dx} = 2\ln{(2)} y$

    Recall that y= 4^x so we get

    $\displaystyle \frac{dy}{dx} = 2\ln{(2)}*4^x$

    In general: $\displaystyle \frac{d}{dx}(a^x) = a\ln{(a^x)}$
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  4. #4
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    Quote Originally Posted by mattc View Post
    This is a pretty simple one I guess but I'm not understanding whats happening here two part question:

    First simply finding d/dx(4^x)?

    The ti-89 says 2*ln(2)*4^x and I'm not getting the mechanics of it.
    The rule is

    $\displaystyle \frac d{dx}\left[a^x\right]=(\ln a)a^x.$

    To see why this is so, you may rewrite $\displaystyle a^x$ as $\displaystyle e^{x\ln a}$ and apply the more familiar rule for differentiating $\displaystyle e^x.$
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  5. #5
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    $\displaystyle y = a^x
    $

    $\displaystyle \ln{y} = \ln{a^x}$

    $\displaystyle \ln{y} = x \cdot \ln{a}$

    $\displaystyle \frac{d}{dx}[\ln{y} = x \cdot \ln{a}]$

    $\displaystyle \frac{y'}{y} = \ln{a}$

    $\displaystyle y' = \ln{a} \cdot y = \ln{a} \cdot a^x $
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  6. #6
    Super Member redsoxfan325's Avatar
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    With regard to your second question: $\displaystyle \frac{d}{dx}\frac{4^xx^{1/3}}{tan(x)}$, you want to use the product and quotient rules.

    $\displaystyle \frac{d}{dx} = \frac{tan(x)*\frac{d}{dx}[4^xx^{1/3}]-4^xx^{1/3}*\frac{d}{dx}[tan(x)]}{tan^2(x)}$

    $\displaystyle \frac{d}{dx}[4^xx^{1/3}] = 4^x*\frac{1}{3}x^{-\frac{2}{3}}+x^{\frac{1}{3}}*ln(4)*4^x$

    $\displaystyle \frac{d}{dx}[tan(x)] = sec^2(x)$

    Plugging this all yields: $\displaystyle \frac{d}{dx} = \frac{tan(x)*(4^x*\frac{1}{3}x^{-\frac{2}{3}}+x^{\frac{1}{3}}*ln(4)*4^x)-4^xx^{1/3}*sec^2(x)}{tan^2(x)}$

    I'll leave the simplification to you.
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  7. #7
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    wow for some reason I had to keep staring at it but it sunk in.

    Thanks to all!
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