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Thread: Derivatives of natural logarithms

  1. #1
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    Derivatives of natural logarithms

    I am just learning how to do these problems. I have to find the derivative of y with respect to t for

    y= ln (t^2)

    Can someone explain the formula for doing a problem like this?
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  2. #2
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    Quote Originally Posted by KarlosK View Post
    I am just learning how to do these problems. I have to find the derivative of y with respect to t for

    y= ln (t^2)

    Can someone explain the formula for doing a problem like this?
    You are supposed to know that the derivative of $\displaystyle f(x)=\ln(x)$ is $\displaystyle f'(x)=\frac1x$.

    Since you don't have a single variable t you have to use

    a) some laws of logaritm: $\displaystyle y = \ln(t^2)~\implies~y=2\ln(t)~\implies~ y'=\frac2t$

    b) the chain rule: $\displaystyle y = \ln(t^2)~\implies~y'=\frac1{t^2} \cdot 2t ~\implies~ y'=\frac2t$

    Be aware that the logrithm function is only defined for positive arguments.
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    Quote Originally Posted by KarlosK View Post
    I am just learning how to do these problems. I have to find the derivative of y with respect to t for

    y= ln (t^2)

    Can someone explain the formula for doing a problem like this?
    Dear KarlosK,

    You have to use,

    1) $\displaystyle \frac{d}{dx}lnx=\frac{1}{x}$

    2)The Chain rule

    Do you have any questions?? Please don't hesitate to ask!!
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    Quote Originally Posted by earboth View Post
    You are supposed to know that the derivative of $\displaystyle f(x)=\ln(x)$ is $\displaystyle f'(x)=\frac1x$.

    Since you don't have a single variable t you have to use

    a) some laws of logaritm: $\displaystyle y = \ln(t^2)~\implies~y=2\ln(t)~\implies~ y'=\frac2t$

    b) the chain rule: $\displaystyle y = \ln(t^2)~\implies~y'=\frac1{t^2} \cdot 2t ~\implies~ y'=\frac2t$

    Be aware that the logrithm function is only defined for positive arguments.
    In part a, I am having trouble seeing how you went from y=2 ln(t) to the y'= 2/t can you just explain that one step.

    Also do you need to do both parts a/b, or is that just a way to check the answer? Thanks for the help!!
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    Quote Originally Posted by KarlosK View Post
    In part a, I am having trouble seeing how you went from y=2 ln(t) to the y'= 2/t can you just explain that one step.

    Also do you need to do both parts a/b, or is that just a way to check the answer? Thanks for the help!!
    Dear KarlosK,

    Do you know that, $\displaystyle \frac{d}{dx}lnx=\frac{1}{x}$ Thats what he had used when going from y=2 ln(t) to the y'= 2/t.

    a) and b) are two methods of doing the same problem. You can do it anyway you wish.

    Hope this helps.
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    Quote Originally Posted by KarlosK View Post
    In part a, I am having trouble seeing how you went from y=2 ln(t) to the y'= 2/t can you just explain that one step.

    Also do you need to do both parts a/b, or is that just a way to check the answer? Thanks for the help!!
    Hi KarlosK, so taking the derivative of a natural logarithmic function is simply $\displaystyle \frac{1}{x} $. In your case your $\displaystyle x $ would be simply $\displaystyle t^2 $. By chain rule, after taking the derivative of the first function, theres is a function within a function which is the $\displaystyle t^2 $, so there for it becomes $\displaystyle 2t $. Yet, yielding the final solution as $\displaystyle y' = \frac{2}{t} $.
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