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Differential of 2^x

  1. Feb 23, 2006 #1
    Just came across a question today with 2^x and realised i didnt know how to differentiate it. The entire function i had to differentiate was
    [math]y = 2^x + x -4[math]

    I tried taking logs but couldnt get anywhere near the true answer.

    What is the correct method for this?

    Thank's alot!
     
  2. jcsd
  3. Feb 23, 2006 #2

    arildno

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    Rewrite [itex]2^{x}=e^{xlog(2)}[/itex] and use the chain rule on your right-hand side in order to find an expression for [itex]\frac{d}{dx}2^{x}[/itex]
     
  4. Feb 23, 2006 #3
    Do you know the rule for the derivative of an expression [tex] a^x [/tex]?
     
  5. Feb 23, 2006 #4
    y = a^u

    y' = lna * a^u * u'

    y = 2^x y' = ln2 * 2^x * 1
    derivative of (x-4) is 1

    so the answer is y' = 2^x * ln2 + 1
     
  6. Feb 24, 2006 #5

    VietDao29

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    DO NOT EVER POST COMPLETE SOLUTION. THE OP NEEDS TO GIVE IT A TRY HIMSELF!!!!!!!!:grumpy: :grumpy: :grumpy: :grumpy: :grumpy:
    --------------------
    mr bob, just try what arildno suggested you.
    [tex]2^{x}=e^{xlog(2)}[/tex]
    Now differentiate both sides with respect to x, we have:
    [tex](2^{x})'=(e^{xlog(2)})' = ?[/tex]
    --------------------
    Or one can try another way:
    Let:
    [tex]y = 2 ^ x[/tex]
    Take log of both sides:
    [tex]\ln y = \ln (2 ^ x) = x \ln 2[/tex]
    Now differentiate both sides wth respect to x gives:
    [tex](\ln y)' = (x \ln 2)'[/tex]
    [tex]\Leftrightarrow \frac{y'_x}{y} = \ln 2[/tex]
    [tex]\Leftrightarrow y'_x = y \ln 2 = 2 ^ x \star ln 2[/tex]
    Now let's try the first way to see if you arrive at the same answer.
    From there, can you solve your problem? :)
     
  7. Feb 27, 2006 #6
    that is wrong!

    sorry to say so but you see the obvious right there!
    please correct. thanks
     
  8. Feb 27, 2006 #7

    What are you talking about? I can't see anything wrong with that...
     
  9. Feb 27, 2006 #8
    sorry because i don't understand the term "+1" in the result y'
    please explain .
     
  10. Feb 27, 2006 #9
    Did you look at the original post? Look at the funtion that we are trying to differentiate here and maybe you'll figure it out.
     
  11. Feb 28, 2006 #10
    ok i agree with you!
    sorry aabout that.
    i want to ask a question if you can solve?
    y= ksin2x.tan3x all over 3cos4x with x is an accute angle and k is a constant.
    can you have the general derivative and the value of that function at the point which is the inteception of y with y =-x and the value of k is 1:)
    this is not a challenge but a hard question to ask you.
    this is my apology
    see ya!
     
  12. Feb 28, 2006 #11
    I'm sorry, but I can't make heads or tails of what you just said, I get the part where you say what the function is but what is that about a general derivative and a point which is an intersection of something?
     
  13. Feb 28, 2006 #12

    HallsofIvy

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    First, it's a really bad idea to "hijack" someone elses thread to ask your onw question- start your own thread. Second, even without using LaTex you ought to be able to write that more clearly. I think you mean
    y= ksin(2x)tan(3x)/(3cos(4x)).

    If you know anything at all about calculus you should see that neither k nor the fact that "x is an acute angle" are relevant. Yes, it certainly has a "general derivative"- its tedious but just apply the product, quotient, and chain rules.

    As for solving x= sin(2x)tan(3x)/(3cos(4x)), that's much harder. In general equations like that cannot be solved algebraically. I would recommend using Newton's method to get a numerical solution.
     
  14. Feb 28, 2006 #13
    i am sorry.
    let take my apology.
    actually i just plan to play with you.
    i am a jounor in high school.
    i don't want this will be a mathematical fight between us.
    please forgive me for that.
    can you give me the result for the Newton's method. i really don't know about that method.
    thanks!
     
  15. Mar 1, 2006 #14

    VietDao29

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    Here's an article on Newton's method.
    It state that:
    [tex]x_{n + 1} = x_n - \frac{f(x_n)}{f'(x_n)}[/tex]
    Now the solution to the equation f(x) = 0 is given by:
    [tex]x = \lim_{n \rightarrow \infty} x_n[/tex]
    There's also an example in the article, just apply it here, and see what you get.
    Can you go from here? :)
     
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