Calculate the derivative (explicitly) [on hold]











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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







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put on hold as off-topic by user302797, Chinnapparaj R, Leucippus, José Carlos Santos, Cesareo yesterday


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." – user302797, 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




    Would you know how to calculate it if the limits were $[0, e^x]$ instead of $[e^{-x}, e^x]$?
    – Connor Harris
    2 days ago






  • 1




    Any tips? Yes: chain rule. Answer $(e^x+e^{-x})sqrt{1+x^2}$.
    – GEdgar
    2 days ago

















up vote
0
down vote

favorite
1












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









New contributor




Juliana de Souza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











put on hold as off-topic by user302797, Chinnapparaj R, Leucippus, José Carlos Santos, Cesareo yesterday


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." – user302797, 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




    Would you know how to calculate it if the limits were $[0, e^x]$ instead of $[e^{-x}, e^x]$?
    – Connor Harris
    2 days ago






  • 1




    Any tips? Yes: chain rule. Answer $(e^x+e^{-x})sqrt{1+x^2}$.
    – GEdgar
    2 days ago















up vote
0
down vote

favorite
1









up vote
0
down vote

favorite
1






1





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









New contributor




Juliana de Souza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











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









New contributor




Juliana de Souza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Juliana de Souza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited yesterday





















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Juliana de Souza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 2 days ago









Juliana de Souza

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636




New contributor




Juliana de Souza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Juliana de Souza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Juliana de Souza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




put on hold as off-topic by user302797, Chinnapparaj R, Leucippus, José Carlos Santos, Cesareo yesterday


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." – user302797, 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.




put on hold as off-topic by user302797, Chinnapparaj R, Leucippus, José Carlos Santos, Cesareo yesterday


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." – user302797, 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




    Would you know how to calculate it if the limits were $[0, e^x]$ instead of $[e^{-x}, e^x]$?
    – Connor Harris
    2 days ago






  • 1




    Any tips? Yes: chain rule. Answer $(e^x+e^{-x})sqrt{1+x^2}$.
    – GEdgar
    2 days ago
















  • 2




    Would you know how to calculate it if the limits were $[0, e^x]$ instead of $[e^{-x}, e^x]$?
    – Connor Harris
    2 days ago






  • 1




    Any tips? Yes: chain rule. Answer $(e^x+e^{-x})sqrt{1+x^2}$.
    – GEdgar
    2 days ago










2




2




Would you know how to calculate it if the limits were $[0, e^x]$ instead of $[e^{-x}, e^x]$?
– Connor Harris
2 days ago




Would you know how to calculate it if the limits were $[0, e^x]$ instead of $[e^{-x}, e^x]$?
– Connor Harris
2 days ago




1




1




Any tips? Yes: chain rule. Answer $(e^x+e^{-x})sqrt{1+x^2}$.
– GEdgar
2 days ago






Any tips? Yes: chain rule. Answer $(e^x+e^{-x})sqrt{1+x^2}$.
– GEdgar
2 days ago












1 Answer
1






active

oldest

votes

















up vote
4
down vote



accepted










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



















  • 1




    I think $g(e^{-x})=sqrt{1+log^2(e^{-x})} =sqrt{1+(-x)^2} = sqrt{1+x^2}$
    – GEdgar
    2 days ago












  • @GEdgar Sorry, yes.
    – J.G.
    2 days ago










  • @GEdgar Yep. Then the result is $sqrt{1+x^2} (e^x+e^{-x})$ Thanks a lot.
    – Juliana de Souza
    2 days ago




















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
4
down vote



accepted










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



















  • 1




    I think $g(e^{-x})=sqrt{1+log^2(e^{-x})} =sqrt{1+(-x)^2} = sqrt{1+x^2}$
    – GEdgar
    2 days ago












  • @GEdgar Sorry, yes.
    – J.G.
    2 days ago










  • @GEdgar Yep. Then the result is $sqrt{1+x^2} (e^x+e^{-x})$ Thanks a lot.
    – Juliana de Souza
    2 days ago

















up vote
4
down vote



accepted










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



















  • 1




    I think $g(e^{-x})=sqrt{1+log^2(e^{-x})} =sqrt{1+(-x)^2} = sqrt{1+x^2}$
    – GEdgar
    2 days ago












  • @GEdgar Sorry, yes.
    – J.G.
    2 days ago










  • @GEdgar Yep. Then the result is $sqrt{1+x^2} (e^x+e^{-x})$ Thanks a lot.
    – Juliana de Souza
    2 days ago















up vote
4
down vote



accepted







up vote
4
down vote



accepted






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














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 2 days ago

























answered 2 days ago









J.G.

19.7k21932




19.7k21932








  • 1




    I think $g(e^{-x})=sqrt{1+log^2(e^{-x})} =sqrt{1+(-x)^2} = sqrt{1+x^2}$
    – GEdgar
    2 days ago












  • @GEdgar Sorry, yes.
    – J.G.
    2 days ago










  • @GEdgar Yep. Then the result is $sqrt{1+x^2} (e^x+e^{-x})$ Thanks a lot.
    – Juliana de Souza
    2 days ago
















  • 1




    I think $g(e^{-x})=sqrt{1+log^2(e^{-x})} =sqrt{1+(-x)^2} = sqrt{1+x^2}$
    – GEdgar
    2 days ago












  • @GEdgar Sorry, yes.
    – J.G.
    2 days ago










  • @GEdgar Yep. Then the result is $sqrt{1+x^2} (e^x+e^{-x})$ Thanks a lot.
    – Juliana de Souza
    2 days ago










1




1




I think $g(e^{-x})=sqrt{1+log^2(e^{-x})} =sqrt{1+(-x)^2} = sqrt{1+x^2}$
– GEdgar
2 days ago






I think $g(e^{-x})=sqrt{1+log^2(e^{-x})} =sqrt{1+(-x)^2} = sqrt{1+x^2}$
– GEdgar
2 days ago














@GEdgar Sorry, yes.
– J.G.
2 days ago




@GEdgar Sorry, yes.
– J.G.
2 days ago












@GEdgar Yep. Then the result is $sqrt{1+x^2} (e^x+e^{-x})$ Thanks a lot.
– Juliana de Souza
2 days ago






@GEdgar Yep. Then the result is $sqrt{1+x^2} (e^x+e^{-x})$ Thanks a lot.
– Juliana de Souza
2 days ago





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