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 $\displaystyle \frac{13\sqrt{10}-2}{6}$.