Question

If h(x)=integral (x^2 to 2) sqr root (1+t^4)dt

Answer 1

\[\Large\rm h(x)=\int\limits_{x^2}^2 \sqrt{1+t^4}~dt\]Do we need to find h'(x)?

Answer 2

my guess, which certainly could be wrong, is that you are asked for the derivative of this thing
like i said, i could be wrong

Answer 3

@zepdrix guessed it too

Answer 4

heh :p

Answer 5

Sorry, I forgot to ask the question. I am looking for h'(x).

Answer 6

\[\Large h(x)=\int\limits_{x^2}^2 \sqrt{1+t^4}~dt\]
\[=\Large h(x)=-\int\limits_{2}^{x^2}\sqrt{1+t^4}~dt\]
since the derivative of the integral is the integrand, replace \(t\) by \(x^2\) and then via the chain rule, multiply the result by \(2x\)