Sorry, it looks a little confusing now that I've written it out.
It should be an integral with an upper bound of 1 and a lower of x of √(1+t^2)
Let f(x) = int 1 x √1 + t^2 dt
Use your calculator to find f’(1)
2 answers
If you mean
f(x) = ∫[x,1] √(1+t^2) dt
then
f'(x) = -√(1+x^2)
so, f'(1) = -√2
This is just the chain rule in reverse. If
F(t) = ∫f(t) dt
then
F'(t) = f(t)
So, if u,v are functions of x, then
∫[u,v] f(t) dt = F(t) [u,v] = F(v)-F(u)
Now take derivatives with the chain rule, and you get
d/dx (∫[u,v] f(t) dt) = f(v) dv/dx - f(u) du/dx
f(x) = ∫[x,1] √(1+t^2) dt
then
f'(x) = -√(1+x^2)
so, f'(1) = -√2
This is just the chain rule in reverse. If
F(t) = ∫f(t) dt
then
F'(t) = f(t)
So, if u,v are functions of x, then
∫[u,v] f(t) dt = F(t) [u,v] = F(v)-F(u)
Now take derivatives with the chain rule, and you get
d/dx (∫[u,v] f(t) dt) = f(v) dv/dx - f(u) du/dx
1 answer