How do you calculate the derivative of #intsqrt(3t+ sqrt(t)) dt# from #[5, tanx]#?

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Answer 1 · 2015-05-26T19:41:19

Fundamental Theorem of Calculus, Part 1.

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

This question is the chain rule version:

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

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

#d/dx int_5^(tanx) f(t) dt = f(tanx) * sec^2x#

So for this problems we get:

#sqrt(3tanx+sqrt(tanx))* sec^2 x#, or, written more clearly:

# sec^2 x sqrt(3tanx+sqrt(tanx))#