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Stupid question

  1. Sep 29, 2010 #1
    I am currently learning calculus and just had my lecture on the chain rule.

    I noticed that we haven't learned how to take the derivative of a function like 2^2+x or 3^4+x.
    Any example works.. Is this something I will learn later as I progress through calculus or what?
     
  2. jcsd
  3. Sep 30, 2010 #2

    Mark44

    Staff: Mentor

    Assuming you meant what you wrote, these are very simple to differentiate. 2^2 + x = 4 + x, and its derivative is 1.

    3^4 + x = 81 + x, and its derivative is also 1.

    Now, assuming you meant 2^(2 + x) = 22 + x, and 3^(4 + x) = 34 + x, these functions can be differentiated when you know the chain rule form of the derivative of the exponential function.
     
  4. Sep 30, 2010 #3
    chain rule : F'(x) = f(g(x)) g'(x)

    example: take f(x) = 2sin(x) ..............f'(x)=2sin(x)cos(x)
     
  5. Sep 30, 2010 #4

    phyzguy

    User Avatar
    Science Advisor

    Not quite right. The derivative of 2^x is not just 2^x, you need to proceed as follows:
    [tex]\frac{d}{dx} 2^x = \frac{d}{dx} e^{x log2} = \frac{d}{dx} (e^x)^{log2} = e^x log2*(e^x)^{log2-1} = log2*2^x [/tex]

    So:
    [tex]\frac{d}{dx} 2^{sin(x)} = log2*cos(x)*2^{sin(x)}[/tex]
     
  6. Sep 30, 2010 #5
    my bad. not even a good example.
     
  7. Sep 30, 2010 #6

    HallsofIvy

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    Science Advisor

    Generally, for any positive a, the derivative of [itex]a^x[/itex], with respect to x, is [itex](ln a)a^x[/itex]. Of course, if a= e, ln(a)= ln(e)= 1.
     
  8. Sep 30, 2010 #7
    So finish this off ....
    Derivative of 2^(2 +x) = (Ln 2) [2^(2+x)] (2+x)' = (Ln 2) [2^(2+x)]
    and
    Derivative of 3^(4 +x) = (Ln 3) [3^(4+x)] (4+x)' = (Ln 3) [3^(4+x)]

    Just for fun checkout Derivative of 2^(2+x^2)
    http://www.wolframalpha.com/input/?i=derivative+2^%282+%2B+x^2%29+dx
    Click on "Show Steps" to see the solution
     
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