Question

Calculate d/dt when a=t^2, b=2 and sqrt(x+1)dx.

My answer turns out to be incorrect despite my checking it several times. I got -2/3t* sqrt(t^2+1)

Can someone explain to me what is wrong with my answer and how to go about obtaining it the right way? Thank you everyone. :-)

Answer 1

f(t) = ∫[2,t^2] √(x+1) dx
df/dt = √(t^2+1) (2t)

Recall that d/dt ∫[a,b(t)] f(x) dx = f(b(t)) db/dt

see wikipedia's article on differentiating under the integral sign.

Answer 2

Thank you for the equation and the useful article. :)

So this is what I get.

f(t) dt= 2(0)- [sqrt(t^2+1) * 2t]
f(t) dt= -sqrt(t^2+1) *2t


Is this correct? Thank you so much. :-)