Question

the derivative of

Answer 1

\[\int\limits_{0}^{x^2}\sin(t^3)dt)=? \]

Answer 2

please and thank you :D

Answer 3

\[2xsinx^6\]

Answer 4

By the fundamental theorem of calculus,
\[ F(b) - F(a) = \int^b_a f(t) \ dt \]
where F' = f.

Now, use that and the chain rule to find your derivative. I.e., let
\[ G(x) = \int_0^{x^2} \sin t^3 \ dt \]
then G(x) = F(x^2) - F(0) where F' = sin x^3. Thus
\[ G'(x) = \frac{d \ }{dx} (F(x^2) - F(0) ) = \ ... \]

Answer 5

grate work