Calculate the derivative (explicitly) [closed]












0












$begingroup$


How to explicitly calculate this derivative? Any tips?



$$frac{d}{dx}biggl[int_{e^{-x}}^{e^x}sqrt{1+(ln t)^2},dtbiggr]$$



I tried to solve it in several ways, including computer software but I could not!






calculus integration derivatives







share|cite|improve this question











$endgroup$



closed as off-topic by Saad, Chinnapparaj R, Leucippus, José Carlos Santos, Cesareo Dec 1 '18 at 10:07


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, Chinnapparaj R, Leucippus, José Carlos Santos, Cesareo

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 2




    $begingroup$
    Would you know how to calculate it if the limits were $[0, e^x]$ instead of $[e^{-x}, e^x]$?
    $endgroup$
    – Connor Harris
    Nov 30 '18 at 17:55






  • 1




    $begingroup$
    Any tips? Yes: chain rule. Answer $(e^x+e^{-x})sqrt{1+x^2}$.
    $endgroup$
    – GEdgar
    Nov 30 '18 at 17:57


















0












$begingroup$


How to explicitly calculate this derivative? Any tips?



$$frac{d}{dx}biggl[int_{e^{-x}}^{e^x}sqrt{1+(ln t)^2},dtbiggr]$$



I tried to solve it in several ways, including computer software but I could not!






calculus integration derivatives







share|cite|improve this question











$endgroup$



closed as off-topic by Saad, Chinnapparaj R, Leucippus, José Carlos Santos, Cesareo Dec 1 '18 at 10:07


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, Chinnapparaj R, Leucippus, José Carlos Santos, Cesareo

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 2




    $begingroup$
    Would you know how to calculate it if the limits were $[0, e^x]$ instead of $[e^{-x}, e^x]$?
    $endgroup$
    – Connor Harris
    Nov 30 '18 at 17:55






  • 1




    $begingroup$
    Any tips? Yes: chain rule. Answer $(e^x+e^{-x})sqrt{1+x^2}$.
    $endgroup$
    – GEdgar
    Nov 30 '18 at 17:57
















0












0








0


1



$begingroup$


How to explicitly calculate this derivative? Any tips?



$$frac{d}{dx}biggl[int_{e^{-x}}^{e^x}sqrt{1+(ln t)^2},dtbiggr]$$



I tried to solve it in several ways, including computer software but I could not!






calculus integration derivatives







share|cite|improve this question











$endgroup$




How to explicitly calculate this derivative? Any tips?



$$frac{d}{dx}biggl[int_{e^{-x}}^{e^x}sqrt{1+(ln t)^2},dtbiggr]$$



I tried to solve it in several ways, including computer software but I could not!






calculus integration derivatives




calculus integration derivatives






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 1 '18 at 19:32







Juliana de Souza

















asked Nov 30 '18 at 17:52









Juliana de SouzaJuliana de Souza

656




656




closed as off-topic by Saad, Chinnapparaj R, Leucippus, José Carlos Santos, Cesareo Dec 1 '18 at 10:07


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, Chinnapparaj R, Leucippus, José Carlos Santos, Cesareo

If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by Saad, Chinnapparaj R, Leucippus, José Carlos Santos, Cesareo Dec 1 '18 at 10:07


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, Chinnapparaj R, Leucippus, José Carlos Santos, Cesareo

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 2




    $begingroup$
    Would you know how to calculate it if the limits were $[0, e^x]$ instead of $[e^{-x}, e^x]$?
    $endgroup$
    – Connor Harris
    Nov 30 '18 at 17:55






  • 1




    $begingroup$
    Any tips? Yes: chain rule. Answer $(e^x+e^{-x})sqrt{1+x^2}$.
    $endgroup$
    – GEdgar
    Nov 30 '18 at 17:57
















  • 2




    $begingroup$
    Would you know how to calculate it if the limits were $[0, e^x]$ instead of $[e^{-x}, e^x]$?
    $endgroup$
    – Connor Harris
    Nov 30 '18 at 17:55






  • 1




    $begingroup$
    Any tips? Yes: chain rule. Answer $(e^x+e^{-x})sqrt{1+x^2}$.
    $endgroup$
    – GEdgar
    Nov 30 '18 at 17:57










2




2




$begingroup$
Would you know how to calculate it if the limits were $[0, e^x]$ instead of $[e^{-x}, e^x]$?
$endgroup$
– Connor Harris
Nov 30 '18 at 17:55




$begingroup$
Would you know how to calculate it if the limits were $[0, e^x]$ instead of $[e^{-x}, e^x]$?
$endgroup$
– Connor Harris
Nov 30 '18 at 17:55




1




1




$begingroup$
Any tips? Yes: chain rule. Answer $(e^x+e^{-x})sqrt{1+x^2}$.
$endgroup$
– GEdgar
Nov 30 '18 at 17:57






$begingroup$
Any tips? Yes: chain rule. Answer $(e^x+e^{-x})sqrt{1+x^2}$.
$endgroup$
– GEdgar
Nov 30 '18 at 17:57












1 Answer
1






active

oldest

votes


















4












$begingroup$

Note that $frac{d}{dx}int_c^{f(x)}g(t)dt=f'(x)g(f(x))$ by the chain rule, so $$frac{d}{dx}int_{e^{-x}}^{e^x}g(t)dt=e^xg(e^x)+e^{-x}g(e^{-x}).$$For your case, $g(x)=sqrt{1+ln^2 x}$, so the result is $$e^xsqrt{1+x^2}+e^{-x}sqrt{1+x^2}=(e^x+e^{-x})sqrt{1+x^2}.$$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    I think $g(e^{-x})=sqrt{1+log^2(e^{-x})} =sqrt{1+(-x)^2} = sqrt{1+x^2}$
    $endgroup$
    – GEdgar
    Nov 30 '18 at 18:03












  • $begingroup$
    @GEdgar Sorry, yes.
    $endgroup$
    – J.G.
    Nov 30 '18 at 18:05










  • $begingroup$
    @GEdgar Yep. Then the result is $sqrt{1+x^2} (e^x+e^{-x})$ Thanks a lot.
    $endgroup$
    – Juliana de Souza
    Nov 30 '18 at 18:09




















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

Note that $frac{d}{dx}int_c^{f(x)}g(t)dt=f'(x)g(f(x))$ by the chain rule, so $$frac{d}{dx}int_{e^{-x}}^{e^x}g(t)dt=e^xg(e^x)+e^{-x}g(e^{-x}).$$For your case, $g(x)=sqrt{1+ln^2 x}$, so the result is $$e^xsqrt{1+x^2}+e^{-x}sqrt{1+x^2}=(e^x+e^{-x})sqrt{1+x^2}.$$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    I think $g(e^{-x})=sqrt{1+log^2(e^{-x})} =sqrt{1+(-x)^2} = sqrt{1+x^2}$
    $endgroup$
    – GEdgar
    Nov 30 '18 at 18:03












  • $begingroup$
    @GEdgar Sorry, yes.
    $endgroup$
    – J.G.
    Nov 30 '18 at 18:05










  • $begingroup$
    @GEdgar Yep. Then the result is $sqrt{1+x^2} (e^x+e^{-x})$ Thanks a lot.
    $endgroup$
    – Juliana de Souza
    Nov 30 '18 at 18:09


















4












$begingroup$

Note that $frac{d}{dx}int_c^{f(x)}g(t)dt=f'(x)g(f(x))$ by the chain rule, so $$frac{d}{dx}int_{e^{-x}}^{e^x}g(t)dt=e^xg(e^x)+e^{-x}g(e^{-x}).$$For your case, $g(x)=sqrt{1+ln^2 x}$, so the result is $$e^xsqrt{1+x^2}+e^{-x}sqrt{1+x^2}=(e^x+e^{-x})sqrt{1+x^2}.$$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    I think $g(e^{-x})=sqrt{1+log^2(e^{-x})} =sqrt{1+(-x)^2} = sqrt{1+x^2}$
    $endgroup$
    – GEdgar
    Nov 30 '18 at 18:03












  • $begingroup$
    @GEdgar Sorry, yes.
    $endgroup$
    – J.G.
    Nov 30 '18 at 18:05










  • $begingroup$
    @GEdgar Yep. Then the result is $sqrt{1+x^2} (e^x+e^{-x})$ Thanks a lot.
    $endgroup$
    – Juliana de Souza
    Nov 30 '18 at 18:09
















4












4








4





$begingroup$

Note that $frac{d}{dx}int_c^{f(x)}g(t)dt=f'(x)g(f(x))$ by the chain rule, so $$frac{d}{dx}int_{e^{-x}}^{e^x}g(t)dt=e^xg(e^x)+e^{-x}g(e^{-x}).$$For your case, $g(x)=sqrt{1+ln^2 x}$, so the result is $$e^xsqrt{1+x^2}+e^{-x}sqrt{1+x^2}=(e^x+e^{-x})sqrt{1+x^2}.$$






share|cite|improve this answer











$endgroup$



Note that $frac{d}{dx}int_c^{f(x)}g(t)dt=f'(x)g(f(x))$ by the chain rule, so $$frac{d}{dx}int_{e^{-x}}^{e^x}g(t)dt=e^xg(e^x)+e^{-x}g(e^{-x}).$$For your case, $g(x)=sqrt{1+ln^2 x}$, so the result is $$e^xsqrt{1+x^2}+e^{-x}sqrt{1+x^2}=(e^x+e^{-x})sqrt{1+x^2}.$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 30 '18 at 18:05

























answered Nov 30 '18 at 17:57









J.G.J.G.

23.6k22338




23.6k22338








  • 1




    $begingroup$
    I think $g(e^{-x})=sqrt{1+log^2(e^{-x})} =sqrt{1+(-x)^2} = sqrt{1+x^2}$
    $endgroup$
    – GEdgar
    Nov 30 '18 at 18:03












  • $begingroup$
    @GEdgar Sorry, yes.
    $endgroup$
    – J.G.
    Nov 30 '18 at 18:05










  • $begingroup$
    @GEdgar Yep. Then the result is $sqrt{1+x^2} (e^x+e^{-x})$ Thanks a lot.
    $endgroup$
    – Juliana de Souza
    Nov 30 '18 at 18:09
















  • 1




    $begingroup$
    I think $g(e^{-x})=sqrt{1+log^2(e^{-x})} =sqrt{1+(-x)^2} = sqrt{1+x^2}$
    $endgroup$
    – GEdgar
    Nov 30 '18 at 18:03












  • $begingroup$
    @GEdgar Sorry, yes.
    $endgroup$
    – J.G.
    Nov 30 '18 at 18:05










  • $begingroup$
    @GEdgar Yep. Then the result is $sqrt{1+x^2} (e^x+e^{-x})$ Thanks a lot.
    $endgroup$
    – Juliana de Souza
    Nov 30 '18 at 18:09










1




1




$begingroup$
I think $g(e^{-x})=sqrt{1+log^2(e^{-x})} =sqrt{1+(-x)^2} = sqrt{1+x^2}$
$endgroup$
– GEdgar
Nov 30 '18 at 18:03






$begingroup$
I think $g(e^{-x})=sqrt{1+log^2(e^{-x})} =sqrt{1+(-x)^2} = sqrt{1+x^2}$
$endgroup$
– GEdgar
Nov 30 '18 at 18:03














$begingroup$
@GEdgar Sorry, yes.
$endgroup$
– J.G.
Nov 30 '18 at 18:05




$begingroup$
@GEdgar Sorry, yes.
$endgroup$
– J.G.
Nov 30 '18 at 18:05












$begingroup$
@GEdgar Yep. Then the result is $sqrt{1+x^2} (e^x+e^{-x})$ Thanks a lot.
$endgroup$
– Juliana de Souza
Nov 30 '18 at 18:09






$begingroup$
@GEdgar Yep. Then the result is $sqrt{1+x^2} (e^x+e^{-x})$ Thanks a lot.
$endgroup$
– Juliana de Souza
Nov 30 '18 at 18:09





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