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differentiate 2^x

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    how would you differentiate 2^x
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    2^x = e^xln2
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    (Original post by uxa595)
    how would you differentiate 2^x
    Remember differentiating,  y =a^x , gives,  \displaystyle \frac{dy}{dx} = a^x lna
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    (Original post by Coulson72)
    2^x = e^xln2
    I got this but i don't think it's right

    (Original post by raheem94)
    Remember differentiating,  y =a^x , gives,  \displaystyle \frac{dy}{dx} = a^x lna
    How do you know a^x foes to a^x lna?
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    (Original post by nm786)
    remeber: dy/dx=x^n=nx^n-1
    2^x=(1)2x^(1-1)=2*1=2
    No.
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    (Original post by uxa595)
    I got this but i don't think it's right



    How do you know a^x foes to a^x lna?
    Its a formula.

    If  y= a^x
    Take the natural log of both sides.
     lny = lna^x \implies lny=xlna
    Differentiate implicitly,
      \displaystyle \frac1y \frac{dy}{dx} = lna \implies \frac{dy}{dx} = ylna

    We know,  y=a^x , hence,  \displaystyle \frac{dy}{dx} = ylna \implies \frac{dy}{dx} = a^x lna

    Do you get it?
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    (Original post by uxa595)
    I got this but i don't think it's right



    How do you know a^x foes to a^x lna?
    If y= a^x

    ln(y)= xln(a)

    1/y(dy/dx)= ln(a)

    dy/dx= a^x(ln(a))
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    (Original post by nm786)
    is this for c1?
    no
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    (Original post by f1mad)
    If y= a^x

    ln(y)= xln(a)

    1/y(dy/dx)= ln(a)

    dy/dx= a^x(ln(a))
    My is better
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    I got this:

    y=2^x
    therefore
    y= e^(xln2)
    y= e^x . e^ln2
    y= 2e^x

    is that right?
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    (Original post by nm786)
    is this for c1?
    C1? Holy **** I haven't done this for C4.
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    (Original post by uxa595)
    I got this:

    y=2^x
    therefore
    y= e^(xln2)
    y= e^x . e^ln2
    y= 2e^x

    is that right?
    No,

    Example:

     \displaystyle y=3^x = e^{ln3^x} = e^{xln3} \\ \frac{dy}{dx} = ln3 \times  e^{xln3} = ln3 \times e^{ln3^x} = 3^xln3

    This could have directly been found using the formula i gave you.
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    (Original post by raheem94)
    Its a formula.

    If  y= a^x
    Take the natural log of both sides.
     lny = lna^x \implies lny=xlna
    Differentiate implicitly,
      \displaystyle \frac1y \frac{dy}{dx} = lna \implies \frac{dy}{dx} = ylna

    We know,  y=a^x , hence,  \displaystyle \frac{dy}{dx} = ylna \implies \frac{dy}{dx} = a^x lna

    Do you get it?
    Or... (if you're not fond of implicit differentiation):

    \displaystyle y=a^x \implies y=e^{\ln a^x} = e^{x\ln a}

    \displaystyle \implies \frac{dy}{dx} = \ln a \ e^{x\ln a} = a^x \ln a
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    (Original post by uxa595)
    I got this:

    y=2^x
    therefore
    y= e^(xln2)
    y= e^x . e^ln2
    y= 2e^x

    is that right?
    No.

    If y= e^(xln2)

    dy/dx= ln2*e^(xln2)

    And you should remember: e^(xln2)= 2^x.
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    (Original post by notnek)
    Or... (if you're not fond of implicit differentiation):

    \displaystyle y=a^x \implies y=e^{\ln a^x} = e^{x\ln a}

    \displaystyle \implies \frac{dy}{dx} = \ln a \ e^{x\ln a} = a^x \ln a
    See post#14 in this thread, i know it mate.
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    (Original post by raheem94)
    No,

    Example:

     \displaystyle y=3^x = e^{ln3^x} = e^{xln3} \\ \frac{dy}{dx} = ln3 \times  e^{xln3} = ln3 \times e^{ln3^x} = 3^xln3

    This could have directly been found using the formula i gave you.
    Ahhh right, thanks, i've got it now

    I can't remember doing this ever in class
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    (Original post by raheem94)
    See post#14 in this thread, i know it mate.
    I assumed you knew it. I posted it for the OP.
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    Just tried this myself

    Doing it implicitly seems the most straight forward

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