How do you use the Fundamental Theorem of Calculus to find the derivative of #int sqrt(6t + sqrt (t)) dt# from 1 to tanx?

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Answer 1 · 2016-08-14T19:51:10

# =sec^2 sqrt(6 tan x + sqrt (tan x)) x#

FTC part 1 tells us that ;

#d/(du) int_a^u f(t) \ dt = f(u)#

and by the chain rule, if #u = u(x)# then

#d/(dx)( int_a^(u(x)) f(t) \ dt )#

#d/(du)( int_a^(u(x)) f(t) \ dt ) * (du)/(dx)#

#= f(u) (du)/dx#

so #d/dx (int_1^(tan x) sqrt(6t + sqrt (t)) dt) #

#= sqrt(6 tan x + sqrt (tan x)) d/dx (tan x)#

# =sec^2 sqrt(6 tan x + sqrt (tan x)) x#