How do you use the Fundamental Theorem of Calculus to find the derivative of #int sqrt(3t+ sqrt(t)) dt# from 5 to tanx?

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Answer 1 · 2018-04-28T09:05:04

#= sec^2 x \ sqrt(3tan x + sqrt(tan x)) #

FTC, Pt 1: if #F(u)= int _a^u \ f(t)\ dt #, then #F'(u)=f(u) #.

With u(x) = tan x, and #F = F(u(x))#, it follows from the chain rule that:

#(dF)/(dx) = F'(u) \ u'(x)#

So:

#d/(dx) ( int_5^(tan(x)) sqrt(3t+ sqrt(t)) \ \ dt) #

#= sqrt(3tan x + sqrt(tan x)) \ d/(dx) (tan x ) #

#= sec^2 x \ sqrt(3tan x + sqrt(tan x)) #