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Thread: chain rule issue with derivative involving ln and e

  1. #1
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    chain rule issue with derivative involving ln and e

    y = ln [(e^x + e^-x) / 2] We need to find the derivative of this.

    So I extract the 1/2 so it becomes

    y = 1/2 ln (e^x + e^-x)

    I'm not sure how the chain rule is applied here. The final answer is; (e^x - e^-x) / (e^x + e^-x)
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Archduke01 View Post
    y = ln [(e^x + e^-x) / 2] We need to find the derivative of this.

    So I extract the 1/2 so it becomes

    y = 1/2 ln (e^x + e^-x)

    I'm not sure how the chain rule is applied here. The final answer is; (e^x - e^-x) / (e^x + e^-x)
    "extract" the 1/2?? And exactly what rule says we can do that?!

    $\displaystyle y = \ln \frac {e^x + e^{-x}}2$

    $\displaystyle = \ln (e^x + e^{-x}) - \ln 2$

    Now differentiate using the rule: $\displaystyle \frac d{dx} \ln f(x) = \frac {f'(x)}{f(x)}$
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    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Archduke01 View Post
    y = ln [(e^x + e^-x) / 2] We need to find the derivative of this.

    So I extract the 1/2 so it becomes

    y = 1/2 ln (e^x + e^-x)

    I'm not sure how the chain rule is applied here. The final answer is; (e^x - e^-x) / (e^x + e^-x)
    You can't 'extract' the 2. It is associated with the Logarithm. You must be careful

    $\displaystyle y=\ln(e^x-e^{-x})-\ln2$

    Now

    Let $\displaystyle u=e^x-e^{-x}$

    Then

    $\displaystyle y'=\frac{u'}{u}=\frac{e^x+e^{-x}}{e^x-e^{-x}}=\coth{x}$
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by VonNemo19 View Post
    You can't 'extract' the 2. It is associated with the Logarithm. You must be careful

    $\displaystyle y=\ln(e^x-e^{-x})-\ln2$

    Now

    Let $\displaystyle u=e^x-e^{-x}$

    Then

    $\displaystyle y'=\frac{u'}{u}=\frac{e^x+e^{-x}}{e^x-e^{-x}}=\coth{x}$
    It's $\displaystyle e^x + e^{-x}$

    @OP: If we wanted hyperbolic functions for the answer, we could have replaced $\displaystyle \frac {e^x + e^{-x}}2$ with $\displaystyle \cosh x$ from the get go
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    Quote Originally Posted by Jhevon View Post
    "extract" the 1/2?? And exactly what rule says we can do that?!

    $\displaystyle y = \ln \frac {e^x + e^{-x}}2$

    $\displaystyle = \ln (e^x + e^{-x}) - \ln 2$

    Now differentiate using the rule: $\displaystyle \frac d{dx} \ln f(x) = \frac {f'(x)}{f(x)}$
    if c is a constant then d ca/dx = c * d a/dx

    But my badness, seems like I misunderstood the rule.

    I don't understand the coth x cosh x stuff; I don't think we've learned that yet.
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  6. #6
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Jhevon View Post
    It's $\displaystyle e^x + e^{-x}$ Oops

    @OP: If we wanted hyperbolic functions for the answer, we could have replaced $\displaystyle \frac {e^x + e^{-x}}2$ with $\displaystyle \cosh x$ from the get go I know this. I was trying to get a "Hey, what is coth,?" out of the OP
    Thanks for the correction.
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    $\displaystyle = \ln (e^x + e^{-x}) - \ln 2
    $

    how did ln 1/2 become - ln 2?
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  8. #8
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Archduke01 View Post
    $\displaystyle = \ln (e^x + e^{-x}) - \ln 2$$\displaystyle
    $

    how did ln 1/2 become - ln 2?
    From the basic log rule $\displaystyle \ln\frac{a}{b}=\ln{a}-\ln{b}.$
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    Quote Originally Posted by VonNemo19 View Post
    From the basic log rule $\displaystyle \ln\frac{a}{b}=\ln{a}-\ln{b}.$
    Thanks bromosapien.

    $\displaystyle y'=\frac{u'}{u}=\frac{e^x+e^{-x}}{e^x-e^{-x}}=\coth{x}
    $

    Is this from applying the chain rule?

    How did you guys get rid of the - ln 2? I assume you derived it, getting - 1/2 but at that point I'm stumped.
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  10. #10
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Archduke01 View Post
    Thanks bromosapien.

    $\displaystyle y'=\frac{u'}{u}=\frac{e^x+e^{-x}}{e^x-e^{-x}}=\coth{x}
    $

    Is this from applying the chain rule?

    How did you guys get rid of the - ln 2? I assume you derived it, getting - 1/2 but at that point I'm stumped.
    ln(2) is a constant....
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  11. #11
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    Quote Originally Posted by Jhevon View Post
    ln(2) is a constant....
    Ah yes of course, I was just -

    Thanks for the help, gentlemen. Sorry for the whole back and forth.
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  12. #12
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Archduke01 View Post
    Thanks bromosapien.

    $\displaystyle y'=\frac{u'}{u}=\frac{e^x+e^{-x}}{e^x-e^{-x}}=\coth{x}$$\displaystyle
    $

    Is this from applying the chain rule?

    How did you guys get rid of the - ln 2? I assume you derived it, getting - 1/2 but at that point I'm stumped.
    $\displaystyle \ln2$ is a constant. Constants dissapear when differentiating.

    Regarding the chain rule...


    You know that $\displaystyle \frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$ right?

    Well, like I said before (if you forget about that nasty sign mistake that I made), let $\displaystyle u=e^x+e^{-x}$, then by the chain rule we have

    $\displaystyle \frac{dy}{dx}=\frac{dy}{du}{\frac{du}{dx}}=\frac{1 }{u}\frac{du}{dx}=\frac{1}{e^x+e^{-x}}\cdot(e^x-e^{-x})$.

    See what I'm sayin' brominator?
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