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Math Help - Need help on this question. thanks guys

  1. #1
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    Exclamation Need help on this question. thanks guys

    (i) Find the numerical value of the derivative of 3^x when x=2.
    The answer given is 9.89

    How do i solve this?

    My attempt:
    Firstly, i equate y=3^x
    then i tried using ln to solve but i got stuck somewhere..
    lny=xln3
    ...

    Please help me! Thank you!
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  2. #2
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    Re: Need help on this question. thanks guys

    its okay guys. I managed to solve it (finally!)

    Heres my answer:

    Let y=3^x
    y=3^x
    lny=ln3^x (ln both sides)
    lny=xln3 (due to the law of log: lnx^a = alnx)
    y= e^xln3 (change to index form)
    dy/dx= e^xln3(ln3) (differentiate y)
    dy/dx = 9.89 (3 s.f.)
    Thanks from topsquark
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  3. #3
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    Re: Need help on this question. thanks guys

    Another way to do this: y= 3^x so ln(y)= x ln(3). Differentiate both sides with respect to x:
    \frac{1}{y}\frac{dy}{dx}= ln(3) so \frac{dy}{dx}= y ln(3)= 3^x ln(3).

    Going back to the definition of derivative, the difference quotient would be \frac{3^{x+h}- 3^x}{h}= \frac{3^x3^h- 3^x}{h}= 3^x\frac{3^h- 1}{h}. So the derivative is given by \lim_{h\to 0} 3^x \frac{3^h- 1}{h}=  3^x\left(\lim_{h\to 0}\frac{3^h- 1}{h}\right)

    Notice that the first part of that, 3^x does not have an "h" while the limit does not involve "x". That means that the derivative of 3^x is just 3^x times a constant. (Of course there is nothing special about "3"- exactly the same thing can be done with an positive number "a", though the constant varies with a.) Using 3^x= e^{ln(3^x)}= e^{xln(3)} shows that the constant is ln(3).
    Thanks from pumbaa213
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