Calculate the derivative (explicitly) [closed]
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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
asked Nov 30 '18 at 17:52
Juliana de SouzaJuliana de Souza
656
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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.
add a comment |
$begingroup$
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
asked Nov 30 '18 at 17:52
Juliana de SouzaJuliana de Souza
656
$endgroup$
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
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Would you know how to calculate it if the limits were $[0, e^x]$ instead of $[e^{-x}, e^x]$?
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– Connor Harris
Nov 30 '18 at 17:55
1
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Any tips? Yes: chain rule. Answer $(e^x+e^{-x})sqrt{1+x^2}$.
$endgroup$
– GEdgar
Nov 30 '18 at 17:57
add a comment |
$begingroup$
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
asked Nov 30 '18 at 17:52
Juliana de SouzaJuliana de Souza
656
$endgroup$
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
asked Nov 30 '18 at 17:52
Juliana de SouzaJuliana de Souza
656
asked Nov 30 '18 at 17:52
Juliana de SouzaJuliana de Souza
656
edited Dec 1 '18 at 19:32
Juliana de Souza
asked Nov 30 '18 at 17:52
Juliana de SouzaJuliana de Souza
656
asked Nov 30 '18 at 17:52
Juliana de SouzaJuliana de Souza
656
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
add a comment |
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
add a comment |
1 Answer
1
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oldest
votes
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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 $$e^xsqrt{1+x^2}+e^{-x}sqrt{1+x^2}=(e^x+e^{-x})sqrt{1+x^2}.$$
answered Nov 30 '18 at 17:57
J.G.J.G.
23.6k22338
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1
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I think $g(e^{-x})=sqrt{1+log^2(e^{-x})} =sqrt{1+(-x)^2} = sqrt{1+x^2}$
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– GEdgar
Nov 30 '18 at 18:03
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@GEdgar Sorry, yes.
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– J.G.
Nov 30 '18 at 18:05
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@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
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $$e^xsqrt{1+x^2}+e^{-x}sqrt{1+x^2}=(e^x+e^{-x})sqrt{1+x^2}.$$
answered Nov 30 '18 at 17:57
J.G.J.G.
23.6k22338
$endgroup$
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
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@GEdgar Sorry, yes.
$endgroup$
– J.G.
Nov 30 '18 at 18:05
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@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
add a comment |
$begingroup$
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 $$e^xsqrt{1+x^2}+e^{-x}sqrt{1+x^2}=(e^x+e^{-x})sqrt{1+x^2}.$$
answered Nov 30 '18 at 17:57
J.G.J.G.
23.6k22338
$endgroup$
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
add a comment |
$begingroup$
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 $$e^xsqrt{1+x^2}+e^{-x}sqrt{1+x^2}=(e^x+e^{-x})sqrt{1+x^2}.$$
answered Nov 30 '18 at 17:57
J.G.J.G.
23.6k22338
$endgroup$
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 $$e^xsqrt{1+x^2}+e^{-x}sqrt{1+x^2}=(e^x+e^{-x})sqrt{1+x^2}.$$
answered Nov 30 '18 at 17:57
J.G.J.G.
23.6k22338
edited Nov 30 '18 at 18:05
answered Nov 30 '18 at 17:57
J.G.J.G.
23.6k22338
answered Nov 30 '18 at 17:57
J.G.J.G.
23.6k22338
answered Nov 30 '18 at 17:57
J.G.J.G.
23.6k22338
23.6k22338
1
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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
add a comment |
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
add a comment |
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