Question

here is a nice question that i'm working on right now : find derivative of :
\[f(x) = \int\limits_{x}^{2x}\sin(t^2)dt \]

Answer 1

it's pretty easy. you should use the following formula -
\[\int\limits_{\alpha}^{\beta} f(x) = f(\beta) f'(\beta) - f(\alpha)f'(\alpha)\]
So your answer should be -
\[2\sin (4x^2) - \sin(x^2)\]

Answer 2

x appaers in both limits so we can make that :
\[\int\limits_{x}^{2x}\sin(t^2)dt =\int\limits_{x}^{0}\sin(t^2)dt + \int\limits_{0}^{2x}\sin(t^2)dt \]

Answer 3

Yup^^Leibniz One

Answer 4

\[d/dx(-\int\limits_{0}^{x}\sin(t^2)dt) = -\sin(x^2)\]

Answer 5

u = 2x \[y = \int\limits_{0}^{u}\sin(t^2)dt\]
so that
\[dy/dx = (dydu) / (dudx) = ( \sin(u^2)) (2) = 2\sin(4x^2) \]

Answer 6

yup same thing Lfc :D: D

Answer 7

2 sin(4x2) - sin(x2)