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    Hello , I was a bit stuck on this and thought I would ask for help before starting to break out in tears. thanks for all the help in advance!


    how do we differentiate 5^x. in Core 4, one of the question gave the sub (for integration ) to be u=5^x .. But in C3 we never learned any rules as to how to differntiate this ....then how can they expect us to know how to diff this....thanks
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    diffing a^x is always ln(a)a^x where 'a' is a constant. so try do it now.
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    (Original post by cooldudeman)
    diffing a^x is always ln(a)a^x where 'a' is a constant. so try do it now.
    thanks very much ...i never knew this rule..was it present in c3?
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    (Original post by laurawoods)
    thanks very much ...i never knew this rule..was it present in c3?
    nope its in c4 only. remember, sometimes you'd have to apply the chain rule for ones like diffing y=2^(2x), it would be
    2ln(2)2^(2x).
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    (Original post by cooldudeman)
    nope its in c4 only. remember, sometimes you'd have to apply the chain rule for ones like diffing y=2^(2x), it would be
    2ln(2)2^(2x).
    hello but i havent seen this formula actually being mentioned anywhere in the C4 EDEXCEL book or in the formual sheet ....
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    (Original post by laurawoods)
    C4 EDEXCEL book
    Page 50 point 4

    (green/blue book)
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    (Original post by laurawoods)


    Hello , I was a bit stuck on this and thought I would ask for help before starting to break out in tears. thanks for all the help in advance!


    how do we differentiate 5^x. in Core 4, one of the question gave the sub (for integration ) to be u=5^x .. But in C3 we never learned any rules as to how to differntiate this ....then how can they expect us to know how to diff this....thanks
    If y=a^x, then lny=xlna. Differentiating both sides gives \frac{1}{y}.\frac{dy}{dx}=lna
    Multiplying by y then gives \frac{dy}{dx}=ylna

    And so as y=a^x, \frac{dy}{dx}=a^xlna.
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    (Original post by TenOfThem)
    Page 50 point 4

    (green/blue book)
    i am doing a green book and btw pg 50 is on vectors.
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    (Original post by laurawoods)
    i am doing a green book and btw pg 50 is on vectors.
    Well, I cannot quote from the out of date text
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    (Original post by brittanna)
    If y=a^x, then lny=xlna. Differentiating both sides gives \frac{1}{y}.\frac{dy}{dx}=lna
    Multiplying by y then gives \frac{dy}{dx}=ylna

    And so as y=a^x, \frac{dy}{dx}=a^xlna.
    hello there , thanks you are pure genius!!!
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    Also, without using implicit, a^x=e^(xlna) also differentiates into lna(a^x).
 
 
 
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