How do you use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function of #y=int sqrt(2t+sqrt(t))dt# from 5 to #tanx#?

Question

Christopher Agresta · Accepted Answer

To find the derivative of the function of #y=int_5^tanx sqrt(2t+sqrt(t))dt#, we have to pair Part 1 of the Fundamental Theorem of Calculus with the Chain Rule.

FTC Part I tells us that for #g(x)= int_a^x f(t) dt#, the derivative with respect to #x# is #g'(x) = f(x)#

Now we combine this with the chain rule:

For the function #g(u)= int_a^u f(t) dt#, the derivative with respect to #x# is

#d/dx(g(u)) = g'(u) (du)/dx = f(u) (du)/dx#.

So here's the answer to this question:

For #y=int_5^tanx sqrt(2t+sqrt(t))dt#, the derivative with respect to #x# is

#dy/dx = sqrt(2tanx +sqrt(tanx)) * sec^2x#.

Charles Anctil · Answer

undefined To use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function y = ∫sqrt(2t + sqrt(t)) dt from 5 to tan(x), first find the antiderivative of the integrand, which is denoted as capital F(t). Then, evaluate F(tan(x)) - F(5) and take the derivative with respect to x. This gives the derivative of the function with respect to x.