In this question we have to find the derivative of F(x) which is
[tex] F(x)=\int_{x^3}^{2}sin(t^4)dt [/tex]
To find the derivative, our first step is to change the limits of the integral and on doing so , we have to put a minus sign , that is
[tex] F(x)= - \int_{2}^{x^3}sin(t^4)dt [/tex]
Now we use fundamental law of calculus by substituting x^3 for t and we have to use chainrule too, that is
[tex] F'(x)= -(x^3)'*sin((x^3)^4) [/tex]
[tex] F'(x)=-3x^2**sin(x^{12}) [/tex]
So the correct option is the last option .
