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differentiating forms of a^x

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    hey guys I'm having trouble understanding how to differentiate a constant to the power of a variable, for example

    y=5^x
    I know
    dy/dx = 5^x ln(5)

    and this is the same for any integer, but I don't understand the mechanism and get stuck when it's something of the form

    5x^x or 5^x^2
  2. Offline

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    see I don't get how this follows :

    y = x^5
    ln Y = X ln (5)

    differntiate both sides,

    = 1/y = (x/5) + ln(5)

    and then i'm confused.
  3. Offline

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    (Original post by kingkongjaffa)
    hey guys I'm having trouble understanding how to differentiate a constant to the power of a variable, for example

    y=5^x
    I know
    dy/dx = 5^x ln(5)

    and this is the same for any integer, but I don't understand the mechanism and get stuck when it's something of the form

    5x^x or 5^x^2
    Take logs and then differentiate.
  4. Offline

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    I meant 5^x not x^5 which of course is 5x^4...
  5. Offline

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    (Original post by steve2005)
    Take logs and then differentiate.
    can you show me I get stuck after taking logs?
  6. Offline

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    Are you taking logs and then using implicit differentiation?

    y=5^{x^2}

    \ln y = x^2 \ln 5

    and so on?
  7. Offline

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    (Original post by Mr M)
    Are you taking logs and then using implicit differentiation?

    y=5^{x^2}

    \ln y = x^2 \ln 5

    and so on?
    yeah i get stuck there the logs isn't a problem
  8. Offline

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    (Original post by kingkongjaffa)
    can you show me I get stuck after taking logs?
    Here's a quicker way to do it.
    1)Note that a^x=(e^{lna})^x=e^{lna\cdot x}
    2)Remember that e^{bx} \Rightarrow \frac{dy}{dx}=be^{bx}
    3) Letting b=ln(a) yields  \frac{dy}{dx}=ln(a) \cdot e^{lna\cdot x}=ln(a)\cdot a^x
  9. Offline

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    (Original post by kingkongjaffa)
    can you show me I get stuck after taking logs?
    As a general rule:

    y = a^x

    ln(y) = ln(a^x)

    ln(y) = x ln(a)

    Now differentiating - the differential of ln(x) is 1/x, so the differential of ln(y) is going to be 1/y times dy/dx:

    1/y * dy/dx = ln(a) <- differentiated the RHS by treating ln(a) as a constant, just as 3x goes to 3 when you diff.

    dy/dx = y ln(a)

    but y = a^x

    so dy/dx = a^x ln(a)
  10. Offline

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    (Original post by kingkongjaffa)
    yeah i get stuck there the logs isn't a problem
    \ln y =x^2 \ln 5

    \frac{1}{y} \frac{dy}{dx} = (2\ln 5) x

    You know what y = ...?
  11. Offline

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    ok i realised what i was doing wrong and understand how to do it in the form a^x without looking at the rule....
  12. Offline

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    y=5^x
    lny=ln(5^x)
    lny=xln5

    differentiate both sides, remembering ln5 is constant
  13. Offline

    ReputationRep:
    How will one apply chain rule when its related to a fraction in a square root?

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