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Math Help - Differentiation using first principles

  1. #1
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    Differentiation using first principles

    Hi

    How do you differentiate using first principles

    eg 3^x

    Any help will be appreciated
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  2. #2
    Senior Member Sambit's Avatar
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    Let y = f(x) = 3^x

    Then \frac{dy}{dx}=\lim_{h\to\0}\frac{f(x+h)-f(x)}{h} = \lim_{h\to\0}\frac{3^{x+h}-3^x}{h}

    = \lim_{h\to\0}\frac{3^x3^h-3^x}{h} = 3^x \lim_{h\to\0}\frac{3^h-1}{h} = 3^x\ln3
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  3. #3
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    Quote Originally Posted by Sambit View Post
    Let y = f(x) = 3^x

    Then \frac{dy}{dx}=\lim_{h\to\0}\frac{f(x+h)-f(x)}{h} = \lim_{h\to\0}\frac{3^{x+h}-3^x}{h}

    = \lim_{h\to\0}\frac{3^x3^h-3^x}{h} = 3^x \lim_{h\to\0}\frac{3^h-1}{h} = 3^x\ln3

    While true is:

    \lim_{h\to\0}\frac{3^h-1}{h} = \ln(3)

    common knowledge?
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  4. #4
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    Quote Originally Posted by nikki1234 View Post
    Hi

    How do you differentiate using first principles

    eg 3^x

    Any help will be appreciated
    Did you just pluck this example out of the air or where you specifically given it and asked to differentiate it from first principles? (Or were you just asked to differentiate it and there was no mention of first principles?)
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  5. #5
    Senior Member Sambit's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    While true is:

    \lim_{h\to\0}\frac{3^h-1}{h} = \ln(3)

    common knowledge?
    Derivative of 3^x is always 3^x\ln3. Where exactly do you find my mistake?
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  6. #6
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    Quote Originally Posted by Sambit View Post
    Derivative of 3^x is always 3^x\ln3. Where exactly do you find my mistake?
    The point being made is that if one already knows that what is there to prove? In other words, have you not used what is to be proved in the proof? Is that really common knowledge?
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  7. #7
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    Quote Originally Posted by Plato View Post
    The point being made is that if one already knows that what is there to prove? In other words, have you not used what is to be proved in the proof? Is that really common knowledge?
    I'd still like to see the response of the OP to post #4. (Many students don't realise that the questions they get given in their textbook etc. are sanitised and cooked to work. A lot of time gets wasetd when it turns out that the posted question has just been plucked out of the air by the student).
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  8. #8
    Senior Member Sambit's Avatar
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    I can't really follow. In my calculus book, derivative of 3^x is calculated from first principle; and it is done in the way I did it here. Is it not right that every algebraic function can be differentiated using first principle?
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  9. #9
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    Quote Originally Posted by Sambit View Post
    I can't really follow. In my calculus book, derivative of 3^x is calculated from first principle; and it is done in the way I did it here. Is it not right that every algebraic function can be differentiated using first principle?
    What they are asking is "How do you know that this limit"

    \lim_{h \to 0}\frac{3^h-1}{h}=\ln(3)

    You can't use L'hosptials rule because we are trying to prove how to take the derivative.
    It is the same for using a Taylor series.
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  10. #10
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    Quote Originally Posted by TheEmptySet View Post
    What they are asking is "How do you know that this limit"

    \lim_{h \to 0}\frac{3^h-1}{h}=\ln(3)
    ..which is equivalent to what you needed to prove in the first place, it is the derivative of the given function at 0.

    CB
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  11. #11
    Senior Member Sambit's Avatar
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    Ok; so you guys are saying I need to prove \lim_{h \to 0}\frac{3^h-1}{h}=\ln(3) first?
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  12. #12
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    Quote Originally Posted by Sambit View Post
    Ok; so you guys are saying I need to prove \lim_{h \to 0}\frac{3^h-1}{h}=\ln(3) first?
    Here is a suggestion: \frac{3^h-1}{h}=\frac{e^{\ln(3)h}-1}{h}
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  13. #13
    Senior Member Sambit's Avatar
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    Quote Originally Posted by Plato View Post
    Here is a suggestion: \frac{3^h-1}{h}=\frac{e^{\ln(3)h}-1}{h}
    I know the process. i just thought it would not be needed.. Here it goes:

    \frac{3^h-1}{h}=\frac{e^z-1}{\frac{z}{\ln 3}} (where z=h\ln 3)
    =\ln 3\lim_{z\to\0}\frac{e^z-1}{z}=\ln 3


    But now if you say why is \lim_{z\to\0}\frac{e^z-1}{z}=1 ? , I can't answer because I have learned it as a result without proof.
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  14. #14
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    Quote Originally Posted by Sambit View Post
    But now if you say why is \lim_{z\to\0}\frac{e^z-1}{z}=1 ? , I can't answer because I have learned it as a result without proof.
    There are multiple ways that textbook authors present this material. I like the way it is done in Gillman & McDowell.
    But theirs is much different from your question, which is the way James Stewart’s books do it.
    Stewart simply defines \mathbf{e} to the number such that
    \displaystyle\lim _{h \to 0} \frac{{e^h  - 1}}{h} = 1.
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  15. #15
    Grand Panjandrum
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    Quote Originally Posted by Sambit View Post
    I know the process. i just thought it would not be needed.. Here it goes:

    \frac{3^h-1}{h}=\frac{e^z-1}{\frac{z}{\ln 3}} (where z=h\ln 3)
    =\ln 3\lim_{z\to\0}\frac{e^z-1}{z}=\ln 3


    But now if you say why is \lim_{z\to\0}\frac{e^z-1}{z}=1 ? , I can't answer because I have learned it as a result without proof.
    Then it can't be used in a derivation from first principles

    CB
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