How do you use the second fundamental theorem of Calculus to find the derivative of given #int dt/sqrt( 1 + t^2) # from #[x, 2x]#?

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Answer 1 · 2016-07-03T17:15:48

#2/sqrt(1+4x^2) - 1/sqrt(1+x^2)#

#d/dx (int_x^{2x} dt/sqrt( 1 + t^2))#

the FTC part 2 tells us that

if #F(x) = int_a^x quad f(t) quaddt #

then #(dF)/dx = f(x)#

so we first need to arrange the integral so that FTC-2 can be applied

#int_x^{2x} dt/sqrt( 1 + t^2)#

#= color{red}{ int_a^{2x} dt/sqrt( 1 + t^2)} -\color{green}{ int_a^{x} dt/sqrt( 1 + t^2)}#

the green term is now good to go as it matches exactly the patterm given above. we can say that

#d/dx (int_a^{x} dt/sqrt( 1 + t^2)) = 1/sqrt(1+x^2) qquad star#

the term in red needs a little more work

if we re-write it as

# int_a^{color{red}{u}} dt/sqrt( 1 + t^2)# where #u = 2x# then we can again conclude from FTC-2 that

# d/(du) int_a^{color{red}{u}} dt/sqrt( 1 + t^2) = f(u)#

then applying the chain rule

# d/(dx) int_a^{color{red}{u}} dt/sqrt( 1 + t^2) #

# = d/(du) int_a^{color{red}{u}} dt/sqrt( 1 + t^2) * (du)/dx#

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

#= 1/sqrt(1+(2x)^2) * 2 #

combining this with #star# gives

#2/sqrt(1+4x^2) - 1/sqrt(1+x^2)#