How do you calculate the derivative of #int(cos(t^3)+t)# from #x=4# to #x=sinx#?

1 Answer
Aug 10, 2015

Answer:

The derivative is #(cos(sin^{3}(x))+sin(x))*cos(x)#

#=cos(x)cos(sin^{3}(x))+cos(x)sin(x)#

Explanation:

One form of the Fundamental Theorem of Calculus says that if #f# is continuous on some interval #[c,d]# and #F(x)=int_{a}^{x}f(t)\ dt# for some number #a# in #[c,d]#, then #F'(x)=f(x)# for all #x# in #[c,d]#. In other words, you can use the integral in a theoretical/abstract way to "construct" an antiderivative of #f# (that doesn't mean you can find a "nice" formula for #F#...it might not be an "elementary function ").

If, in addition, you have a function #g# that is differentiable on the interval #[c,d]#, and if you create another function #h# by the (composition) formula #h(x)=F(g(x))=int_{a}^{g(x)}f(t)\ dt#, then the Chain Rule implies that #h'(x)=F'(g(x)) * g'(x)=f(g(x)) * g'(x)#.

In the present problem, #f(t)=cos(t^{3})+t# and #g(x)=sin(x)#. Since #g'(x)=cos(x)#, we get the answer:

#h'(x)=d/dx(int_{a}^{sin(x)}(cos(t^{3})+t)\ dt)#

#=(cos(sin^{3}(x))+sin(x))*cos(x)#

#=cos(x)cos(sin^{3}(x))+cos(x)sin(x)#