Question

how to find this derivative,

Answer 1

\[g(x) =\int\limits_{1}^{x} \frac{ 1 }{ t^{3} + 1 }\]

is it:
\[g'(x) = \frac{ d }{ dx }(g'(x) - g'(1)) = g(x) = \frac{ 1 }{ t^{3}+1 }\]

Answer 2

hmm

Answer 3

is the right side integration with respect to dt?

Answer 4

derivative could be undefined...see Newton Leibniz rule for finding derivatives under the integral sign..ur question asks for the derivative of infinity

Answer 5

"is the right side integration with respect to dt?"
yes it is

g(x)=∫1x1t3+1 dt

Answer 6

ok

Answer 7

ya maybe 1/x^3+1

Answer 8

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

Answer 9

hold on...is the upper limit x or infinity??