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Math Help - using Leibniz Rule

  1. #1
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    using Leibniz Rule

    Question. A function is defined by:

    f(t)=\int^t_{\frac{1}{t}}\frac{\sqrt{1+t^2x^2}}{x}  dx.

    Find f\prime(t) and show that f\prime(\sqrt{2})=\frac{13\sqrt{10}-2}{6}.

    I cannot get to the final number... I am making a mistake somewhere (or mistakes!) and I'd appreciate your feedback!

    First, apply Leibniz Rule for differentiation under integral sign:

    f\prime{t}=\int^t_{\frac{1}{t}}\frac{\partial}{\pa  rtial(t)}\frac{\sqrt{1+t^2x^2}}{x}dx +\frac{\sqrt{1+t^2*t^2}}{t} \frac{d}{dt}(t) -\frac{\sqrt{1+t^2*(1/t)^2}}{\frac{1}{t}}\frac{d}{dt}(\frac{1}{t})

    Then I deal with the first integral first: differentiate on t, and then integrate on x (using substitution 1+t^2x^2=u).

    \int^t_{\frac{1}{t}}\frac{\partial}{\partial(t)}\f  rac{\sqrt{1+t^2x^2}}{x}dx=\int^t_{\frac{1}{t}}\fra  c{1}{2x}}\frac{1}{\sqrt{1+t^2x^2}}dx*2tx^2=\int^t_  {\frac{1}{t}}\frac{tx}{\sqrt{1+t^2x^2}}dx=\frac{2\  sqrt{1+t^2x^2}}{t}\mid^t_{\frac{1}{t}}= \frac{\sqrt{1+t^4}}{t}-\frac{\sqrt{2}}{t}

    Adding the other two components from above, I get the final answer
    f\prime(t)=2\frac{\sqrt{1+t^4}}{t}-\frac{\sqrt{2}}{t}+\frac{t\sqrt{2}}{t^2}=2\frac{\s  qrt{1+t^4}}{t}
    and substituting \sqrt{2}, I get f\prime(\sqrt{2})=\sqrt{10}, but not the required \frac{13\sqrt{10}-2}{6}.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Volga View Post
    \frac{\sqrt{1+t^4}}{t}-\frac{\sqrt{2}}{t}

    It should be:

    \dfrac{\sqrt{1+t^4}}{t}+\dfrac{\sqrt{2}}{t}

    Fernando Revilla
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  3. #3
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    Sorry?... Not sure why? and even if I try this sign combo I still cannot get to the final answer... 13 is hard to get, having only 2s )))

    I think I put an extra (unnecessary) 2 in the intermediate differentiation under integral expression - 2s cancel each other out
    \int^t_{\frac{1}{t}}\frac{\partial}{\partial(t)}\f  rac{\sqrt{1+t^2x^2}}{x}dx=\int^t_{\frac{1}{t}}\fra  c{1}{2x}}\frac{1}{\sqrt{1+t^2x^2}}dx*2tx^2=\int^t_  {\frac{1}{t}}\frac{tx}{\sqrt{1+t^2x^2}}dx=\frac{\s  qrt{1+t^2x^2}}{t}\mid^t_{\frac{1}{t}}=\frac{\sqrt{  1+t^4}}{t}-\frac{\sqrt{2}}{t}

    Do you mean here, in the final sum, I need + not -?
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Volga View Post
    Sorry?... Not sure why?
    Sorry, I looked to the complete formula, not to only to the integral (see line 3 below) .

    You have found correctly f'(t) :


    \begin{array}{lcl}<br />
f'(t) =\\<br />
{}\\<br />
\dfrac{d}{dt}\left(\displaystyle\int_{1/t}^t\dfrac{\sqrt{1+t^2x^2}}{x}\;dx\right)= \\<br />
{}\\<br />
\displaystyle\int_{1/t}^t\dfrac{xt}{\sqrt{ 1+t^2x^2 }}\;dx+\dfrac{\sqrt{1+t^4}}{t}+\dfrac{\sqrt{2}}{t}  =\\<br />
{}\\<br />
\dfrac{1}{t}\left[\;\sqrt{1+t^2x^2}\right\;]_{1/t}^{t}+\dfrac{\sqrt{1+t^4}+\sqrt{2}}{t}=\\<br />
{}\\<br />
\dfrac{\sqrt{1+t^4}-\sqrt{2}}{t}+\dfrac{\sqrt{1+t^4}+\sqrt{2}}{t}=\\<br />
{}\\<br />
\dfrac{2\sqrt{1+t^4}}{t}<br /> <br />
\end{array} $<br />

    So,

    \boxed{\;f'(\sqrt{2})=\sqrt{10}\;}

    Fernando Revilla

    P.S. Clearly there is a typo in your book.
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  5. #5
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    I see! Thank you! It's more important for me to know that I apply Leibniz rule correctly than to get the same answer as the textbook.
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  6. #6
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    For good measure, could you please check if the following exercise in using Leibniz Rule for functions involving exponential is correct? Again, I don't have answers in my study guide.

    Question. A function is defined by f(t)=\int^{t^2}_{\frac{1}{t}}\frac{\exp^{-xt}}{x}dx.
    Find f\prime(t)

    Answer.

    f\prime(t)=\int^{t^2}_{\frac{1}{t}}\frac{\partial}  {\partial(t)}\frac{\exp^{-xt}}{x}dx+\frac{-\exp^{-t^2t}}{t^2}d/dt(t^2)-\frac{\exp^{-\frac{1}{t}t}}{\frac{1}{t}}d/dt(\frac{1}{t})= \int^{t^2}_{\frac{1}{t}}\frac{1}{x}\exp^{-xt}(-x)dx+\frac{\exp^{-t^3}}{t^2}2t-\frac{\exp^{-1}}{\frac{1}{t}}(-\frac{1}{t^2})= \frac{1}{t}\exp^{-t^3}-\frac{1}{\exp*t}+2\frac{\exp^{-t^3}}{t}+\frac{1}{\exp*t}= 3\frac{\exp^{-t^3}}{t}.

    Let me know if there are mistakes.
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  7. #7
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Volga View Post
    Thank you!

    More easy: press the Thanks button.


    Quote Originally Posted by Volga View Post
    For good measure, could you please check if the following exercise in using Leibniz Rule for functions involving exponential is correct? Again, I don't have answers in my study guide.

    f(t)=\int^{t^2}_{\frac{1}{t}}\frac{\exp^{-xt}}{x}dx.
    Find f\prime(t)

    Answer ... 3\frac{\exp^{-t^3}}{t}.

    Right.

    Fernando Revilla
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