How do you use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function #y = int(sqrt(2t+sqrt(t))dt)# from 2 to tanx?

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Answer 1 · 2015-04-13T16:52:59

The theorem says:

Given the function:

#y=int_(h(x))^g(x)f(t)dt#

than:

#y'=f(g(x))*g'(x)-f(h(x))*h'(x)#.

So, in our case:

#y'=sqrt(2tanx+sqrttanx)*1/cos^2x-sqrt(2*2+sqrt2*)0rArr#

#y'=sqrt(2tanx+sqrttanx)*1/cos^2x#.

This is because #1/cos^2x# is the derivative of #tanx# and #0# is the derivative of #2#.