# using Leibniz Rule

• Jan 4th 2011, 12:05 AM
Volga
using Leibniz Rule
Question. A function is defined by:

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

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

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:

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

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

Adding the other two components from above, I get the final answer
$\displaystyle f\prime(t)=2\frac{\sqrt{1+t^4}}{t}-\frac{\sqrt{2}}{t}+\frac{t\sqrt{2}}{t^2}=2\frac{\s qrt{1+t^4}}{t}$
and substituting $\displaystyle \sqrt{2}$, I get $\displaystyle f\prime(\sqrt{2})=\sqrt{10}$, but not the required $\displaystyle \frac{13\sqrt{10}-2}{6}$.
• Jan 4th 2011, 02:09 AM
FernandoRevilla
Quote:

Originally Posted by Volga
$\displaystyle \frac{\sqrt{1+t^4}}{t}-\frac{\sqrt{2}}{t}$

It should be:

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

Fernando Revilla
• Jan 4th 2011, 04:37 AM
Volga
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
$\displaystyle \int^t_{\frac{1}{t}}\frac{\partial}{\partial(t)}\f rac{\sqrt{1+t^2x^2}}{x}dx=\int^t_{\frac{1}{t}}\fra c{1}{2x}}\frac{1}{\sqrt{1+t^2x^2}}dx*2tx^2=\int^t_ {\frac{1}{t}}\frac{tx}{\sqrt{1+t^2x^2}}dx=\frac{\s qrt{1+t^2x^2}}{t}\mid^t_{\frac{1}{t}}=\frac{\sqrt{ 1+t^4}}{t}-\frac{\sqrt{2}}{t}$

Do you mean here, in the final sum, I need + not -?
• Jan 4th 2011, 09:56 AM
FernandoRevilla
Quote:

Originally Posted by Volga
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 $\displaystyle f'(t)$ :

$\displaystyle \begin{array}{lcl} f'(t) =\\ {}\\ \dfrac{d}{dt}\left(\displaystyle\int_{1/t}^t\dfrac{\sqrt{1+t^2x^2}}{x}\;dx\right)= \\ {}\\ \displaystyle\int_{1/t}^t\dfrac{xt}{\sqrt{ 1+t^2x^2 }}\;dx+\dfrac{\sqrt{1+t^4}}{t}+\dfrac{\sqrt{2}}{t} =\\ {}\\ \dfrac{1}{t}\left[\;\sqrt{1+t^2x^2}\right\;]_{1/t}^{t}+\dfrac{\sqrt{1+t^4}+\sqrt{2}}{t}=\\ {}\\ \dfrac{\sqrt{1+t^4}-\sqrt{2}}{t}+\dfrac{\sqrt{1+t^4}+\sqrt{2}}{t}=\\ {}\\ \dfrac{2\sqrt{1+t^4}}{t} \end{array}$
$So,$\displaystyle \boxed{\;f'(\sqrt{2})=\sqrt{10}\;}$Fernando Revilla P.S. Clearly there is a typo in your book. • Jan 4th 2011, 12:47 PM Volga I see! Thank you! It's more important for me to know that I apply Leibniz rule correctly than to get the same answer as the textbook. • Jan 4th 2011, 06:08 PM Volga For good measure, could you please check if the following exercise in using Leibniz Rule for functions involving exponential is correct? Again, I don't have answers in my study guide. Question. A function is defined by$\displaystyle f(t)=\int^{t^2}_{\frac{1}{t}}\frac{\exp^{-xt}}{x}dx$. Find$\displaystyle f\prime(t)$Answer.$\displaystyle 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})$=$\displaystyle \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})=$$\displaystyle \frac{1}{t}\exp^{-t^3}-\frac{1}{\exp*t}+2\frac{\exp^{-t^3}}{t}+\frac{1}{\exp*t}=$$\displaystyle 3\frac{\exp^{-t^3}}{t}.$Let me know if there are mistakes. • Jan 4th 2011, 09:39 PM FernandoRevilla Quote: Originally Posted by Volga Thank you! More easy: press the Thanks button. Quote: Originally Posted by Volga For good measure, could you please check if the following exercise in using Leibniz Rule for functions involving exponential is correct? Again, I don't have answers in my study guide.$\displaystyle f(t)=\int^{t^2}_{\frac{1}{t}}\frac{\exp^{-xt}}{x}dx$. Find$\displaystyle f\prime(t)$Answer ...$\displaystyle 3\frac{\exp^{-t^3}}{t}.\$

Right.

Fernando Revilla