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Math Help - logarithmic derivatives

  1. #1
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    logarithmic derivatives

    I've gotten some excellent help on these forums today. A HUGE thank you to everyone who assisted.

    3 more questions giving me issues. If you could at least help get me started I would appreciate that...no need to do all my homework for me lol

    Find the derivative of the following functions-

    f(t)=tan^-1(e^4t-2)tan(e^t+1)




    g(z)=ln(z^2)-5ze^(2z-1)




    Given e^(x^2)+y^6=3xtan(y), find y'


    I'm not really sure where to start on these
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  2. #2
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    Do you remember the Power Rule, the Product Rule, and the Chain Rule? Here are two additional formulas that will help:

    \frac{d}{dx}\ln x=\frac{1}{x}\;\;\;\;\;\;\;\;\;\;\frac{d}{dx}\tan^  {-1}x=\frac{1}{1+x^2}

    For the final problem, we may differentiate both sides and express y' in terms of x and y.
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  3. #3
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    Quote Originally Posted by bassman111 View Post
    I've gotten some excellent help on these forums today. A HUGE thank you to everyone who assisted.

    3 more questions giving me issues. If you could at least help get me started I would appreciate that...no need to do all my homework for me lol

    Find the derivative of the following functions-

    f(t)=tan^-1(e^4t-2)tan(e^t+1)




    g(z)=ln(z^2)-5ze^(2z-1)




    Given e^(x^2)+y^6=3xtan(y), find y'


    I'm not really sure where to start on these
    I can't remember the derivative of a inverse tangent, but just look it up and use the chain rule. For example, to find g'(z), use the chain rule:

    You should know that the derivative of a function  f(u)=ln(u) is just \frac{1}{u}

    So to differentiate g(z):

    g'(z)=\frac{2z}{z^2}-(5e^{2z-1}+10ze^{2z-1})

    =\frac{2}{z}-e^{2z-1}(5+10z)

    Do you see how I applied the power rule and the chain rule?
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  4. #4
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    I understand the product and the power rule...the chain rule still confuses me a bit though.
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  5. #5
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    Quote Originally Posted by bassman111 View Post
    I understand the product and the power rule...the chain rule still confuses me a bit though.
    Ok, if we have the function:

    f(z)=ln(z^2)

    This can be expressed as a composite function. f(u(z))=ln(u) where u=z^2

    \frac{df}{dz}=\frac{df}{du}\frac{du}{dz}

    \frac{df}{du}=\frac{1}{u}=\frac{1}{z^2}

    \frac{du}{dz}=2z

    If we multiply the two derivatives above we have \frac{2z}{z^2}=\frac{2}{z}

    This is how I used the chain rule to get the first term in: =\frac{2}{z}-e^{2z-1}(5+10z)
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  6. #6
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    Quote Originally Posted by adkinsjr View Post
    I can't remember the derivative of a inverse tangent, but just look it up and use the chain rule. For example, to find g'(z), use the chain rule:

    You should know that the derivative of a function  f(u)=ln(u) is just \frac{1}{u}

    So to differentiate g(z):

    g'(z)=\frac{2z}{z^2}-(5e^{2z-1}+10ze^{2z-1})

    =\frac{2}{z}-e^{2z-1}(5+10z)

    Do you see how I applied the power rule and the chain rule?
    so is =\frac{2}{z}-e^{2z-1}(5+10z) the derivative? or do I need to eliminate 'e' from the equation?
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  7. #7
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    Quote Originally Posted by bassman111 View Post
    so is =\frac{2}{z}-e^{2z-1}(5+10z) the derivative? or do I need to eliminate 'e' from the equation?
    Yes, that is the derivative. Whenever you differentiate the exponential function e^u you will get \frac{du}{dz}e^{u} where u=u(z)
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  8. #8
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    ok, so if x=e^4t-2 and y=e^t+1

    I have...
    (1/(1+x))*(sec^2y)
    for the first question. Is this correct so far? And where would I go from here...?
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  9. #9
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    double post
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  10. #10
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    Quote Originally Posted by bassman111 View Post
    ok, so if x=e^4t-2 and y=e^t+1

    I have...
    (1/(1+x))*(sec^2y)
    for the first question. Is this correct so far? And where would I go from here...?
    You have to apply the power rule to differentiate tan^{-1}(x)tan(y), but then you also have to consider that x and y are both composite functions of t, which means you have apply the chain rule to both terms.

    The derivative start as:

    \frac{dx}{dt}\frac{1}{1+x^2}tan(y)+\frac{dy}{dt}ta  n(x)\frac{d}{dy}tan(y)
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  11. #11
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    :\ either my teacher did not do a good job explaining or I'm not very good at this because what you responded with just does not make sense to me...sorry. Things are not looking good for this assignment
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  12. #12
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    Sorry, not "POWER RULE," I meant product rule.
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  13. #13
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    Quote Originally Posted by bassman111 View Post
    :\ either my teacher did not do a good job explaining or I'm not very good at this because what you responded with just does not make sense to me...sorry. Things are not looking good for this assignment
    Calculus isn't easy. It took the genius of Isaac Newton to develop, so don't be discouraged.

    Do you understand why you have to apply the product rule to tan^{-1}(x)tan(y) ?

    Remember that you defined x and y as functions of t when you wrote:

    x=e^{4t-2} and y=e^{t+1}

    This means that tan^{-1}(x) is a function of a function. The same is true with tan(y). In order to differentiate I have to apply the chain rule.

    If we consider the function:

    tan^{-1}(x)

    we have the derivative:

    \frac{dx}{dt}\frac{1}{1+x^2}

    \frac{dx}{dt}=4e^{4t-2}

    4e^{4t-2}{1+x^2}

    You don't want leave it in this form since x=e^{4t-2}, you should substitute this for the x^2

    Can you tell what the derivative with respect to t of tan(y) is?
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  14. #14
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    Quote Originally Posted by adkinsjr View Post
    Calculus isn't easy. It took the genius of Isaac Newton to develop, so don't be discouraged.

    Do you understand why you have to apply the power rule to tan^{-1}(x)tan(y) ?

    Remember that you defined x and y as functions of t when you wrote:

    x=e^{4t-2} and y=e^{t+1}

    This means that tan^{-1}(x) is a function of a function. The same is true with tan(y). In order to differentiate I have to apply the chain rule.

    tan^{-1}(x)

    Then the derivative is:

    \frac{dx}{dt}\frac{1}{1+x^2}

    \frac{dx}{dt}=4e^{4t-2}

    4e^{4t-2}{1+x^2}

    You don't want leave it in this form since x=e^{4t-2}, you should substitute this for the x^2

    Can you tell what the derivative with respect to t of tan(y) is?
    sec^2(e^t+1)
    right?
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  15. #15
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    Quote Originally Posted by bassman111 View Post
    sec^2(e^t+1)
    right?
    Exactly. Now you have the derivatives of tan^{-1}(x) and tan(y)

    So the derivative of the product tan^{-1}(x)tan(y) is given by the product rule. Do you see how to apply the product rule to this?
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