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Math Help - How to diffrentiate an exponential function to the function base ?

  1. #1
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    How to diffrentiate an exponential function to the function base ?

    For example ,
    y = 17(22+x)^(41-x)^30


    I tried to use math code but it didn't show the exponential numbers/function properly

     y = 17(22+x) ^(41-x) ^30
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by Nora View Post
    For example ,
    y = 17(22+x)^(41-x)^30


    I tried to use math code but it didn't show the exponential numbers/function properly

     y = 17(22+x) ^(41-x) ^30
    Consider that
    y = 17(22+x)^{(41-x)^{30}}=17\cdot e^{\ln\left((22+x)^{(41-x)^{30}}\right)}=17\cdot e^{(41-x)^{30}\cdot\ln\big(22+x\big)}
    Now use the chain rule to get the derivative of that last term.
    Last edited by Failure; May 6th 2010 at 02:50 AM. Reason: made a mistake by taking the factor 17 into the exponent: corrected
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  3. #3
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    Thank you could you please tell me why use e^ln(y) ?(i mean the last two steps )
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  4. #4
    Super Member Failure's Avatar
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    Quote Originally Posted by Nora View Post
    Thank you could you please tell me why use e^ln(y) ?(i mean the last two steps )
    Why? - Well, because you do not have a rule to differentiate something like f(x):= u(x)^{v(x)} directly (analogous to the product rule or the quotient rule, that is). Instead you do it like this:

    f'(x)=\left(u(x)^{v(x)}\right)'=\left(e^{v(x)\ln u(x)}\right)'=e^{v(x)\ln u(x)}\cdot\left(v(x)\ln u(x)\right)'

    where the last equality holds because of the chain rule. Now you can apply the product rule (and for \ln u(x) the chain rule again) and are done:
    =u(x)^{v(x)}\cdot \left(v'(x)\ln u(x)+v(x)\cdot\frac{1}{u(x)}\cdot u'(x)\right)
    Last edited by Failure; May 6th 2010 at 02:48 AM.
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  5. #5
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    aha I got it .. So it's a rule like other rules (Exponential , product , log ....from where you can get directly the derivative

    Thank you so much

    I'll solve the equation right now
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  6. #6
    Super Member Failure's Avatar
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    Quote Originally Posted by Nora View Post
    aha I got it .. So it's a rule like other rules (Exponential , product , log ....from where you can get directly the derivative

    Thank you so much

    I'll solve the equation right now
    Um, well, yes you might take it as a rule like the other rules, but you would not want to actually memorize it, like you did (or should do) with the other, much more important rules: first, because it is not used that often, and, second, because it is easier to remember that you can replace y=u(x)^{v(x)} by y=e^{v(x)\ln u(x)} and then apply the rules that you already know about.

    P.S: Please note that I have made a silly mistake in my first reply to your question that I have now corrected: the constant factor 17 in front of the power-term can (and should) be left as a factor in front of the remaining exponential mess...
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  7. #7
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    well , It's already written in my book .. when you've said it's a rule I grabbed my book to look for it then I found it under (exponential function to the base of constant ) and compare it to your explanation it was so helpful
    Thank you

    and yes the number 17 i knew it was mistake because it's constant and can't be included in derivative thank you for clarifying
    you made my calculus life easier
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  8. #8
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    Hello, Nora!

    Use logarithmic differentiation.


    y \;=\; 17(22+x)^{(41-x)^{30}}

    Take logs: . \ln(y) \;=\;\ln\bigg[17(22+x)^{(41-x)^{30}}\bigg] \;=\;(41-x)^{30}\cdot\ln[17(22+x)]<br />

    Differentiate implicitly: . \frac{1}{y}\!\cdot\!y' \;=\;(41-x)^{30}\cdot\frac{17}{17(22+x)} + 30(41-x)^{29}(-1)\cdot\ln[17(22+x)]

    . . \frac{y'}{y} \;=\;\frac{(41-x)^{30}}{x+22} - 30(41-x)^{29}\cdot\ln[17(22+x)] \;=\; . \frac{(41-x)^{29}}{x+22}\bigg[(41-x) - 30(22+x)\ln[17(22+x)]\bigg]

    . . y' \;=\;y\cdot\frac{(41-x)^{29}}{22+x}\bigg[(41-x) - 30(22+x)\ln[17(22+x)]\bigg]

    . . y' \;=\;17(22+x)^{(41-x)^{30}}\cdot\frac{(41-x)^{29}}{22+x} \bigg[(41-x) - 30(22+x)\ln[17(22+x)]\bigg]<br /> <br />


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  9. #9
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    thank you

    i was looking for this problem
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  10. #10
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    With something like " y= f(x)^{g(x)}", there are two simple mistakes we could make:

    1) Treat the exponent, g(x), as a constant and use the "power rule" to get
    " y'= g(x)f(x)^{g(x)-1}f'(x)".

    2) Treat the base, f(x), as a constant and use the "exponential rule" to get " y'= ln(f(x)) f(x)^{g(x)}g'(x).

    The interesting thing is that the correct derivative is the sum of those two errors!

    Taking the logarithm of both sides of y= f(x)^{g(x)}, ln(y)= g(x) ln(f(x)). On the left side, (ln(y))'= \frac{1}{y} y'. On the right, using the product rule, (g(x)ln(f(x))'= g'(x)ln(f(x)+ \frac{g(x)}{f(x)}f'(x).

    Multiplying on both sides of \frac{1}{y}y'= g'(x)ln(f(x))+ \frac{g(x)}{f(x)}f'(x) by y= f(x)^{g(x)}, we get
    y'= ln(f(x))f(x)^{g(x)}g'(x)+ g(x)f(x)^{g(x)- 1} g'(x).
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  11. #11
    Member grgrsanjay's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    With something like " y= f(x)^{g(x)}", there are two simple mistakes we could make:

    1) Treat the exponent, g(x), as a constant and use the "power rule" to get
    " y'= g(x)f(x)^{g(x)-1}f'(x)".

    2) Treat the base, f(x), as a constant and use the "exponential rule" to get " y'= ln(f(x)) f(x)^{g(x)}g'(x).

    The interesting thing is that the correct derivative is the sum of those two errors!

    Taking the logarithm of both sides of y= f(x)^{g(x)}, ln(y)= g(x) ln(f(x)). On the left side, (ln(y))'= \frac{1}{y} y'. On the right, using the product rule, (g(x)ln(f(x))'= g'(x)ln(f(x)+ \frac{g(x)}{f(x)}f'(x).

    Multiplying on both sides of \frac{1}{y}y'= g'(x)ln(f(x))+ \frac{g(x)}{f(x)}f'(x) by y= f(x)^{g(x)}, we get
    y'= ln(f(x))f(x)^{g(x)}g'(x)+ g(x)f(x)^{g(x)- 1} g'(x).
    excellent
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  12. #12
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    Thank you very much for helping me
    Soroban

    HallsofIvy


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