Question

How do you use part 1 of Fundamental Theorem of Calculus to find the derivative of the function `y=int (6+v^2)^10 dv` from sinx to cosx?

Answer 1

`[-(6+cos^2x)^10 sinx - (6+sin^2x)^10 cosx]`

If it is F(v) = `int_sinx^cosx (6+v^2)`dv, it can be written as a sum of two integrals= `int_sinx^0 (6+v^2)dv` +`int_0^cosx (6+v^2)dv`

=`int_0^cosx (6+v^2)dv-int_0^sinx (6+v^2)dv`

F'(v) =`d/dx int_0^cosx (6+v^2)dv -d/dx int_o^sinx (6+v^2)dv`

Now applying the Fundamental Theorem of Calculus, the derivative would be `[-(6+cos^2x)^10 sinx - (6+sin^2x)^10 cosx]`