anonymous
  • anonymous
how to find this derivative,
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[g(x) =\int\limits_{1}^{x} \frac{ 1 }{ t^{3} + 1 }\] is it: \[g'(x) = \frac{ d }{ dx }(g'(x) - g'(1)) = g(x) = \frac{ 1 }{ t^{3}+1 }\]
dan815
  • dan815
hmm
dan815
  • dan815
is the right side integration with respect to dt?

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amriju
  • amriju
derivative could be undefined...see Newton Leibniz rule for finding derivatives under the integral sign..ur question asks for the derivative of infinity
anonymous
  • anonymous
"is the right side integration with respect to dt?" yes it is g(x)=∫1x1t3+1 dt
dan815
  • dan815
ok
dan815
  • dan815
ya maybe 1/x^3+1
amriju
  • amriju
Hey Dan..u can't find derivatives of definite integrals lyk this...there's the law I posted avobe..u can check it out..@dan815
amriju
  • amriju
hold on...is the upper limit x or infinity??
dan815
  • dan815
|dw:1377593547766:dw|
dan815
  • dan815
so that same function that was aplied on t must now be the function being applied on x
amriju
  • amriju
if its x..then the answer is 1/(x^2+1) - 1/2
amriju
  • amriju
sorry only 1/(x^2+1)
dan815
  • dan815
isnt it just 1/x^3+1
dan815
  • dan815
how come it became x^2?
amriju
  • amriju
is it t^2 or t^3 in the ques??
anonymous
  • anonymous
3
amriju
  • amriju
oh then x^3...the fonts are bloody small..anyways if u have to learn this kind of differentiation fo for that law..google it..
dan815
  • dan815
amriju
dan815
  • dan815
press ctrl+ +
dan815
  • dan815
it will zoom in
dan815
  • dan815
and ctrl + - to zoom out
amriju
  • amriju
okk..I didn't know that...thanks @dan815
dan815
  • dan815
no prob
anonymous
  • anonymous
\[g'(x)=\frac{ d }{ dx } [ \int\limits_{x}^{1} h(t) dt] \] \[= \frac{ d }{ dx }[H(x) - H(1)]\] \[=h(x)\] where h(x) is 1/(t^3 +1) and H(x) is the antiderivative of h(x) is this the correct way?
dan815
  • dan815
there is no t in h(x)
dan815
  • dan815
h(x) by definition is a function of x
anonymous
  • anonymous
ah the integration sign, input wrongly, should be start at 1 and end at x
amriju
  • amriju
I don't think so...
anonymous
  • anonymous
there is no t in h(x) sorry, i meant h(t) = 1/(t^3+1) H(t) is the anti derivative of h(t)
dan815
  • dan815
oh ok
dan815
  • dan815
heres a way to get an intuition for it, its not a proof though
dan815
  • dan815
try some easier to integrate functions
anonymous
  • anonymous
ok, i get wat u mean
dan815
  • dan815
|dw:1377594399310:dw|
amriju
  • amriju
http://en.wikipedia.org/wiki/Leibniz_integral_rule check "variable limits form" in this article
dan815
  • dan815
|dw:1377594452051:dw|
amriju
  • amriju
Solve by that method
dan815
  • dan815
what method there
dan815
  • dan815
|dw:1377594812292:dw|
amriju
  • amriju
Did u check the link I gave you??..
dan815
  • dan815
yes.
amriju
  • amriju
so did u find the "variable limits form" section in the article??...This ques falls in that category...U see the formula provided there..solve this ques by that formula
dan815
  • dan815
oh i see okay
dan815
  • dan815
oh thats the similar logic i used
dan815
  • dan815
thats how i was saying h'x = f(x) if h't = f(t)
anonymous
  • anonymous
i am learning this "fundamental theorem calculus", http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus. Just want to make confirm is this is how this theorem solve this kind of problem
amriju
  • amriju
yep...its the one...
anonymous
  • anonymous
\[ \Large F(x)=\int_{a}^{g(x)}f(t)dt\\ \Large F'(x)=f(x)*g'(x) \]
UnkleRhaukus
  • UnkleRhaukus
The Fundamental Theorem of Calculus \[\frac{\mathrm d}{\mathrm dx}\int\limits_{h(x)}^{g(x)}f(t)\mathrm dt=f\big(g(x)\big)\cdot g'(x)-f\big(h(x)\big)\cdot h'(x)\]
UnkleRhaukus
  • UnkleRhaukus
i think you final answer should be in terms of 'x' not 't'

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