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    Hi
    i need help in this question

    How we differentiate y=3^x-1

    I have done like that

    1/y dy /dx = ln(3^x-1)

    1/y dy/dx = (x-1)ln (3)
    dy/dx = y (x-1) ln (3)
    dy/dx = (3^x-1)(x-1)ln(3)


    BUT MARK SCHEME
    shows 3^x-1 (ln(3))


    where I get wrong?
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    gcsemusicsucks
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    (Original post by Qer)
    Hi
    i need help in this question

    How we differentiate y=3^x-1
    y=3^{x-1} or y=3^x-1?


    I have done like that

    1/y dy /dx = \ln (3^x-1)
    1/y dy/dx = (x-1) \ln (3)
    If

     \ln y = (x-1) \ln3

    then

    \dfrac{1}{y} \dfrac{dy}{dx}= \ln 3

    since (x-1) \ln 3=x \ln 3- \ln 3 and the derivative of a constant is 0.
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    (Original post by BobbJo)
    y=3^{x-1} or y=3^x-1?


    If

     \ln y = (x-1) \ln3

    then

    \dfrac{1}{y} \dfrac{dy}{dx}= \ln 3

    since (x-1) \ln 3=x \ln 3- \ln 3 and the derivative of a constant is 0.
    y=3^{x-1}
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    (Original post by Qer)

    1/y dy /dx = ln(3^x-1)
    It looks like you've not differentiated the RHS.
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    Alternative to implicit differentiation you could write 3^{x-1}=e^{(x-1)ln(3)} and use chain rule
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    (Original post by Jarred)
    It looks like you've not differentiated the RHS.
    yeah

    you are right
 
 
 
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