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Derivative question

  1. May 29, 2006 #1

    I'm not sure how to do this. Any tips? How would I treat the power?

    Should be x^3 as the power, not 3x.
  2. jcsd
  3. May 29, 2006 #2


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    For that, its probably easiest to simply multiply out the bracket, which would give you an expression with two terms, then differentiate those two terms separately.
    Last edited: May 29, 2006
  4. May 29, 2006 #3
    His tex code seems to suggest that x^2+1 is raised to x^3. In this case, you can easily solve it by taking the logarithm and then differentiate (I find it simpler that way).
  5. May 29, 2006 #4


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    oh. my mistake then
  6. May 29, 2006 #5


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    Okay, I assume that you mean:
    [tex]y = (x ^ 2 + 1) ^ {x ^ 3}[/tex]
    Ok, I'll give you an example:
    Differentiate xx with respect to x.
    There are 2 ways to solve it.
    The first way is to try to change xx into esomething. Then from there, we can just use the Chain Rule to arrive at what the problem asks.
    [tex]x ^ x = \left( e ^ {ln x} \right) ^ x = e ^ {x \ln x}[/tex]
    Now we have:
    [tex]\left( x ^ x \right)' = \left( e ^ {x \ln x} \right)' = (x \ln x)' e ^ {x \ln x} = (\ln x + 1) e ^ {x \ln x} = x ^ x (\ln x + 1)[/tex]
    The second way is to let y = xx
    Now take the natural log of both sides, we have: ln y = ln(xx) = x ln x.
    Now differentiate both sides with respect to x gives:
    [tex]\frac{y'_x}{y} = \ln x + 1[/tex]
    [tex]\Rightarrow y'_x = y( \ln x + 1)[/tex]
    [tex]\Rightarrow y'_x = x ^ x ( \ln x + 1)[/tex]
    The two methods give the same result. Now can you go from here? :)
  7. May 29, 2006 #6


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    Just to expand a bit on the second way and why you can do that...

    In your original problem you are looking for [tex]\frac{d}{dx}((x ^ 2 + 1) ^ {x ^ 3})[/tex]

    Note that there is no y involved anywhere in the expression, but if you introduce a "dummy variable" y and set it equal to the given expression you get [tex]y=(x ^ 2 + 1) ^ {x ^ 3}[/tex]

    Notice now that [tex]\frac{dy}{dx}=\frac{d}{dx}((x ^ 2 + 1) ^ {x ^ 3})[/tex] In other words, if you can manipulate the new equation ([itex]y=(x ^ 2 + 1) ^ {x ^ 3}[/itex])and solve for [itex]\frac{dy}{dx}[/itex] then you will have solved for the initial derivative.
    Last edited: May 29, 2006
  8. May 29, 2006 #7
    Ah, yes. I had forgotten all about solving these buggers with logarithms. Thank you.
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