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    (Original post by RDKGames)
    Yes it is. Look at all the information, there is one equality that shows they're both the same.
    Ohhh because of a^x = e^{x\ln a}

    But then where is the equality above derived from?
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    (Original post by Philip-flop)
    Ohhh because of a^x = e^{x\ln a}

    But then where is the equality above derived from?
    The fact that e^{\ln(a)}=a and that x\ln(a)=\ln(a^x) - just basic log rules from C2
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    (Original post by Philip-flop)
    Ohhh because of a^x = e^{x\ln a}

    But then where is the equality above derived from?
    Didn't you ask this on TSR before and there were quite a few posts explaining it? Or am I mixing you up with someone else?
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    (Original post by RDKGames)
    The fact that e^{\ln(a)}=a and that x\ln(a)=\ln(a^x) - just basic log rules from C2
    Great now I'm trying to see how a^x = e^{x\ln a} is derived from...

    e^{\ln(a)}=a and x\ln(a)=\ln(a^x)
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    (Original post by notnek)
    Didn't you ask this on TSR before and there were quite a few posts explaining it? Or am I mixing you up with someone else?
    I can't recall asking this before, but knowing me it might have been!

    Is there a video anywhere explaining it?
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    (Original post by notnek)
    Didn't you ask this on TSR before and there were quite a few posts explaining it? Or am I mixing you up with someone else?
    Yeah it was him

    (Original post by Philip-flop)
    Great now I'm trying to see how a^x = e^{x\ln a} is derived from...

    e^{\ln(a)}=a and x\ln(a)=\ln(a^x)
    https://www.thestudentroom.co.uk/sho....php?t=4642162
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    (Original post by Philip-flop)
    I can't recall me asking this before, but knowing me it might have been!

    Is there a video anywhere explaining it?
    Have a look through this thread
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    Oh dear, why do I have the memory of a sieve?! Sorry about this

    (Original post by notnek)
    Have a look through this thread
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    (Original post by Philip-flop)
    Oh dear, why do I have the memory of a sieve?! Sorry about this
    If you're still unsure about it then it would be worth going through that thread and quoting posts that you don't understand.
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    (Original post by Philip-flop)
    Also I have no idea how a^x = e^{x\ln a} is linked to the above Seriously feel like giving up this close to the exams!
    You're focusing on a very small part of C4 that I think most students are unsure of. This is not a very good reason to give up yet
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    (Original post by notnek)
    If you're still unsure about it then it would be worth going through that thread and quoting posts that you don't understand.
    Thank you, I think I managed to learn it again so hopefully it'll stick better the second time round!! Sorry about forgetting about my original post!

    So using the basis of what you said on the other thread I tweaked it a little to make sure I understand it whenever I have to revisit it...

     xln a \times ln e = xln a

    (This is because  ln e = 1)

    Using the log power rule on the x gives...
     ln a^x \times ln e = ln a^x

    Using the log power rule on the lna^x gives...
    ln e^{\ln a^x} = ln a^x

    Then cancelling out the logs aka "Anti-log" gives...
     e^{ln a^x} = a^x \Rightarrow e^{xln a} = a^x
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    wow good luck!!!
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    (Original post by Philip-flop)
    Great now I'm trying to see how a^x = e^{x\ln a} is derived from...

    e^{\ln(a)}=a and x\ln(a)=\ln(a^x)
    What's e^{\ln a^x}?
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    (Original post by Philip-flop)
    I feel so stupid! I spent so long trying to figure this out over something that is actually so so so simple!! Thanks for clearing this up!




    Yeah I'm definitely still trying to get to grips with f'(x)[f(x)]^n = \frac{[f(x)]^{n+1}}{n+1} + c

    It may take me a while
    Sorry if this is obvious at all, but you see that if you differentiate the right side how everything fits together. First bring down the power, this cancels the 1/(n+1).
    Then you decrease the power, giving n.
    Now here's the important bit, the chain rule. f(x)^(n+1) is in essence another function, let's call it g(x). You just differentiated g(x) when you brought down the power and reduced it. Now we have to differentiate the function within the function and multiply it with g'(x). That's where the f'(x) is from.

    I hope this helps. Good luck with nailing c4!
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    (Original post by carpetguy)
    Sorry if this is obvious at all, but you see that if you differentiate the right side how everything fits together. First bring down the power, this cancels the 1/(n+1).
    Then you decrease the power, giving n.
    Now here's the important bit, the chain rule. f(x)^(n+1) is in essence another function, let's call it g(x). You just differentiated g(x) when you brought down the power and reduced it. Now we have to differentiate the function within the function and multiply it with g'(x). That's where the f'(x) is from.

    I hope this helps. Good luck with nailing c4!
    Slightly confusing, but I think I get what you're saying. It's difficult to understand without an actual example though
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    (Original post by Philip-flop)
    Slightly confusing, but I think I get what you're saying. It's difficult to understand without an actual example though
    You've been doing these examples...
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    (Original post by RDKGames)
    You've been doing these examples...
    Yeah but I haven't tried using the fact that  \int f'(x)[f(x)]^n = \frac{[f(x)]^{n+1}}{n+1} + c in reverse
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    (Original post by Philip-flop)
    Yeah but I haven't tried using the fact that  \int f'(x)[f(x)]^n = \frac{[f(x)]^{n+1}}{n+1} + c in reverse
    Yeah you have. It's just the chain rule.
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    (Original post by RDKGames)
    Yeah you have. It's just the chain rule.
    Oh really? Why am I having a hard time recognising it from that formula?
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    (Original post by Philip-flop)
    Oh really? Why am I having a hard time recognising it from that formula?
    Because the constants are in different places I suppose...

    \frac{d}{dx}[f(x)]^n = nf'(x)[f(x)]^{n-1} is the form you know (hopefully). So if n \mapsto (n+1) we have:

    \frac{d}{dx}[f(x)]^{n+1}=(n+1)f'(x)[f(x)]^{n} so dividing both sides by (n+1) gives \frac{1}{n+1}\frac{d}{dx}[f(x)]^{n+1}=f'(x)[f(x)]^{n}

    Integrating both sides gives \frac{1}{n+1}[f(x)]^{n+1} = \int f'(x)[f(x)]^{n} .dx
 
 
 
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